Cls Jeead-15-16 Xi Mat Target-1 Set-2 Chapter-3

Chapter

3

Trigonometric Functions Solutions SECTION - A Objective Type Questions (One option is correct) Elementary Trigonometric Functions 1.

The radian measure of 450° is (1)

5 4

(2)

5 3

(3)

5 2

(4) 5

Sol. Answer (3)

2.

The degree measure of (1) 160°

7 is 6

(2) 210°

(3) 250°

(4) 270°

(3) sin1° = sin1c

(4) sin1 

 sin1c 180

(3) cos1° = cos1c

(4) sin1 

1 sin1c 180

(3) 2

(4) 0

Sol. Answer (2) 3.

Which of the following is correct? (1) sin1° > sin1c

(2) sin1° < sin1c

Sol. Answer (2) 4.

Which of the following is correct? (1) cos1° > cos1c

(2) cos1° < cos1c

Sol. Answer (1) 5.

If sin + cosec = 2, then sin2 + cosec2 is equal to (1) 1

(2) 4

Sol. Answer (3) 6.

If tan  = 3 and  lies in the III quadrant, then the value of sin is (1)

1 10

(2) –

1 10

(3)

3 10

(4)

3 10

Sol. Answer (3) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

48 7.

Trigonometric Functions

If tan  

(1)

Solution of Assignment (Set-2)

4 , then sin is 3

4 4 but not 5 5

(2)

4 4 or 5 5

(3)

4 5 but not 5 4

(4)

4 5 and both 5 4

Sol. Answer (2) Correct choice is (2). Since tan    sin  

8.

4 is negative,  lies either in the II quadrant or in the IV quadrant. Thus 3

4 4 if  lies in the II quadrant and sin    , if  lies in the IV quadrant. 5 5

If A lies in the second quadrant and 3 tan A + 4 = 0, then the value of 2cotA – 5cosA + sinA is equal to (1)

53 10

(2)

23 10

(3)

37 10

(4)

7 10

Sol. Answer (2) 3tanA = – 4

5

4

4 tan A   3

3

 2cotA – 5cosA + sinA

⎛ 3 ⎞ ⎛ 3 ⎞ 4 = 2 ⎜⎝ ⎟⎠  5 ⎜⎝ ⎟⎠  4 5 5

9.

=

6 4 3 4 5

=

30  60  16 46 23   20 20 10

The circular wire of diameter 10 cm is cut and placed along the circumference of a circle of diameter 1 m. The angle subtended by the wire at the centre of the circle is equal to (1)

 4

(2)

 3

(3)

 5

(4)

 10

Sol. Answer (3) Diameter of the wire = 10 cm Length of wire = 10 cm Another circle is of diameter = 1 m = 100 cm  Radius = 50 cm Required angle =

Length of arc 10    Radius 50 5

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10. The value of (secA + tanA – 1) (secA – tanA + 1) – 2tanA is equal to (1) secA

(2) 2secA

(3) 0

(4) 1

(3) 60°, 240°

(4) 120°, 240°

Sol. Answer (3) = (secA + tanA – 1) (secA – tan A + 1) – 2tanA = [secA + (tanA – 1)] [secA – (tan A – 1)] – 2tanA = [sec2A –(tanA – 1)2]– 2tanA 2 2 = sec A  tan A  1  2 tan A  2 tan A

=1–1=0

11.

If cos  

1 and 0    360 , then the values of  are 2

(1) 120°, 210°

(2) 120°, 300°

Sol. Answer (4) If 0 <  < 360°, then cos  

1  2  4 ⇒   and    2 3 3 3 3

 = 120° and 240° 12. The value of sin15° + cos105° is (1) 0

(2) 2sin15°

(3) cos15° + sin15°

(4) sin15° – cos15°

(3) 1

(4) 0

(3) 0

(4) 1

(3) 1

(4) –1

Sol. Answer (1) sin15° + cos105° = sin15° + cos(90° + 15°) = sin15° – sin15° =0 13. If sin1 + sin2 + sin3 = 3, then cos1 + cos2 + cos3 = (1) 3

(2) 2

Sol. Answer (4) sin1 = sin2 = sin3 = 1.  1  2  3 

 2

14. The value of cos10° – sin10° is (1) Positive

(2) Negative

Sol. Answer (1) 15. The value of cos1°.cos2°.....cos179° is (1)

1 2

(2) 0

Sol. Answer (2) cos1°.cos2°……cos179° = 0 (as there will be a term cos 90° in this series whose value is zero) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

50


16. If sec  

(1) 


2  3cot  ⎞ 13 ⎛ , 0    ⎟ , then the value of is 5 ⎜⎝ 2⎠ 4  9 sec 2   1

15 352

(2)

15 352

(3)

30 352

(4)

5 352

Sol. Answer (1) sec  

 13 ,0    5 2

5 23 2  3cot  12  = 12 49 4  9 sec 2   1 5

=

13

12 ( sec 2   1  tan2  )

5

3/4 15  88/5 352

17. The value of (cosecA.cosecB  cot A.cot B )2  (cosecA.cot B  cosecB .cot A)2 is (1) 1

(2) 2

(3) 0

(4) –1

Sol. Answer (1)

(cosecA.cosecB  cot A.cot B )2  (cosecA.cot B  cosecB .cot A)2  cosec 2 A(cosec 2B  cot 2 B )  cot 2 A(cosec 2B  cot 2 B )  cosec 2 A  cot 2 A  1 18. If tan   cot   a, then the value of tan4   cot 4  is equal to (1) a 4  4a 2  2

(2) a 4  4a 2  2

(3) a 4  4a 2  2

(4)  a 4  2a 2  4

Sol. Answer (2) tan   cot   a , tan4   cot 4   ? 4 2 2 2 2 tan4   cot 4  = (tan   cot  )  4 tan .cot .(tan   cot  )  6 tan .cot 

 a 4  4{(tan   cot  )2  2 tan .cot  }  6  a 4  4a 2  2

19. If acos + bsin = 3 and asin – bcos = 4, then a2 + b2 has the value (1) 25

(2) 14

(3) 7

(4) 15

Sol. Answer (1) a cos   b sin   3 a sin   b cos   4

Squaring and adding, a2 + b2 = 25 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



51

20. Suppose that a cos = b and c sin = d, for  and some constants a, b, c, d. Then which of the following is true? (1) a 2c 2  b 2c 2  a 2d 2

(2) a 2d 2  b 2c 2  a 2 c 2

(3) b 2c 2  a 2d 2  a 2 c 2

(4) a 2 b 2  b 2c 2  c 2 d 2

Sol. Answer (1)

a cos   b, c sin   d  cos  

b d ,sin   a c

Squaring and adding,

b2 a2



d2 c2

1

 b 2c 2  a 2d 2  a 2c 2

21. The number of intersecting points on the graph for sin x  (1) 3

(2) 4

x for x  [ , ] is 10

(3) 2

(4) 1

Sol. Answer (1)



sin x 

2

x , x  [ , ] 10

Clearly the total number of points of intersection is 3. 22.

(3  sec 2 x )max  (4  tan2 y )min equals (1) 0

(2) 1

(3) 2

(4) 3

(3) 0

(4) 1

Sol. Answer (3)

(3  sec 2 x )max  (4  tan2 y )min  2  4  2

23. If tan  

(1)

p sin   q cos  p , then the value of is p sin   q cos  q

p2  q 2 p2  q 2

(2)

p2  q 2 p2  q 2


52



Sol. Answer (1)

tan  

p p sin   q cos  , ? q p sin   q cos 

Divide numerator and denominator by cos  p tan   q p 2  q 2  p tan   q p 2  q 2

24. The value of sin1.cos2.tan3.cot 4.sec 5.cosec6 is (1) Positive

(2) Negative

(3) Zero

(4) May be positive and Negative

Sol. Answer (2) sin1 cos 2 tan3 cot 4 sec 5 cosec 6      

= negative 25. Which of the following is correct? (1) tan1 > tan2

(2) tan2 > tan1

(3) sin1 < cos1

(4) cos3 > cos2

Sol. Answer (1) Both the options are correct

2 1

tan 1 > tan 2 (from graph) 26. The perimeter of a certain sector of a circle is equal to half that of the circle of which it is a sector. The circular measure of one angle of the sector is (1) (– 2) radian

(2) (+ 2) radian

(3) radian

(4) (– 3) radian

Sol. Answer (1) 2r  l 

1 2r where r is radius of the circle and l is the length of the arc i.e. l  r  2

2r  r   r

⇒   (  2) radian Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



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27. If , , [0, 2] , then the sum of all possible values of , ,  if

sin   

(1)

1 1 , cos  , tan    3, is 2 2

22 3

21 3

(2)

(3)

20 3

(4) 8

Sol. Answer (1)

sin    cos   

1 2

⇒   

  , 2  4 4

  1 ⇒   ,  2 3 3

tan    3 ⇒    

  , 2  3 3

 Sum of all values = 8 

2 22  3 3

28. The angles of a triangle are in A.P. and the number of degrees in the least to the number of radians in greatest is 60 to . The angles in degree are (1) 60°, 60°, 60°

(2) 30°, 60°, 90°

(3) 45°, 60°, 75°

(4) 15°, 60°, 105°

Sol. Answer (2) Let the angles be   d, ,   d  3  180 ⇒   60

  d 180 60   d    d = 30° Hence the angles are 30°, 60°, 90°. 29. If ABCD is a cyclic quadrilateral such that 12tanA – 5 = 0 and 5cosB+3 = 0, then the quadratic equation whose roots are cosC and tanD is (1) 39 x 2  16 x  48  0

(2) 39 x 2  88 x  48  0

(3) 39 x 2  88 x  48  0

(4) 39 x 2  13 x  46  0

Sol. Answer (1) 12 tan A  5  0 , 5 cos B  3  0

 tan A 

5 3  cos B   12 5

∵ ABCD is a cyclic quadrilateral

 A + C = ,

B+D=

 C = – A,

D=–B

cos C  – cos A = 

12 13

tan D   tan B =

4 3


54



⎛ 12 4 ⎞ ⎛ 12 ⎞ 4  ⎟ x  ⎜ ⎟   0  Equation whose roots are cos C and tan D is x 2  ⎜⎝  ⎝ 13 ⎠ 3 13 3 ⎠ 39 x 2  16 x  48  0

30. If cos   sec   2, [0, 2] , then sin8   cos8  equal to (1) –2

(2) 1

(3) 24

(4) 25

Sol. Answer (2) cos   sec   2 ,   [0, 2 ]

 cos   sec   1 

 sin8   cos8  0  1  1 31. If the maximum value of cos(cosx) is a and minimum value is b, then (1) b = cos a

(2) a = cos b

(3) a = 0

(4) b = 0

(3) 2k

(4)

Sol. Answer (1)

y  cos(cos x ) y max  1  a, when x 

 2

y min  cos1  b when x  0

 b = cos a

32. If

1 sin A  cos A 2sin A is  k , then 1 sin A 1  sin A  cos A

(1)

k 2

(2) k

1 k

Sol. Answer (2)

2sin A 1  sin A  cos A k, ? 1  sin A  cos A 1  sin A k

2sin A.(1  sin A  cos A) (1  sin A)2  cos2 A

 k

2sin A.(1  sin A  cos A) (1  sin A)2  (1  sin2 A)



2 sin A.(1  sin A  cot A) (1  sin A).{(1  sin A)  (1  sin A)}



1  sin A  cos A 1  sin A




55

33. If sin x  sin2 x  1, then the value of cos12 x  3cos10 x  3cos8 x  cos6 x  1 is equal to (1) 0

(2) 1

(3) –1

(4) 2

Sol. Answer (1) sin x  sin2 x  1

 sin x  cos2 x

 cos6 x.(cos6 x  3cos4 x  3cos2 x  1)  1  {cos2 x(1  cos2 x )}3  1  {sin x(1  sin x )}3  1  {sin x  sin2 x }3  1  (1)3  1  0 29

⎛ r

⎞

∑ cos ⎜⎝ 2   ⎟⎠  S1

34. If

29

and

r 15

(1) cot

⎛ r

⎞

∑ sin ⎜⎝ 2   ⎟⎠  S2 ,

then

r 15

(2) tan

S1 equals S2

(3) –cot

(4) –tan

Sol. Answer (1) 29

∑

r 15 29

∑

r 15

⎛ r ⎞ cos ⎜  ⎟  S1 ⎝ 2 ⎠ ⎛ r ⎞  ⎟  S2 sin ⎜ ⎝ 2 ⎠ 29

S1 =

∑

r 15

⎛ r ⎞ cos ⎜  ⎟ ⎝ 2 ⎠

⎛ 15 ⎞ ⎛ 16 ⎞ ⎛ 29 ⎞  cos ⎜  ⎟  cos ⎜  ⎟  ....  cos ⎜  ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠  (sin   cos   sin   cos )  ..... 

(Sum of consecutive terms is zero and these are 15 terms in all) S1 = (sin   cos   sin )

 cos  29

S2 =

∑

r 15

⎛ r ⎞  ⎟ sin ⎜ ⎝ 2 ⎠

= – cos   sin   cos  = sin 

S1

S2



cos  sin  = cot


56



35. The maximum value of cos2 cos(33  )  sin2  sin(45  ) is (1) 1 + sin21

(3) 1 + cos21

(2) 2

(4) cos22

Sol. Answer (1)

cos2  cos(33  )  sin2  sin(45  )  cos2 (  cos )  sin2 (  sin )  cos2 (  cos )  sin2 (sin ) ∵ cos  cos(cos )  1

cos2 1  cos2 (cos )  1 and 0  sin2 (sin )  sin2 1 ∵ cos2 1  cos2 (cos )  sin2 (sin )  1  sin2 1

⎞ ⎛  Maximum value = 1  sin2 1 ⎜⎝ At   ⎟⎠ 2

36. If

sin4 x cos4 x 1   , then 2 3 5

(1) tan2 x 

(3)

1 3

sin8 x cos8 x 1   8 27 125

(2) tan2 x 

1 4

(4) sin2 x 

3 4

Sol. Answer (3)

sin4 x cos4 x 1   2 3 5  5(3tan4x + 2) = 6 sec4x  15tan4x + 10 = 6 + 6tan4 x + 12 tan2x  9tan4x – 12tan2 x + 4 = 0 (3tan2x – 2)2 = 0 2⎞ ⎛ 2 ⎜⎝ tan x  3 ⎟⎠

sin2 x  

2 3 & cos2 x  5 5

5

2 3

sin8 x sin8 x  8 27 

16 81  8  625 27  625



5 1  625 125




57

37. If a sin2  b cos2   m, b sin2  a cos   n, a tan   b tan  , then which of the following is true? (1)

1 1 1 1    n m a b

(2)

1 1 1 1    n m a b

2 2 2 2 (3) m  n  a  b

(4) m 2  n 2  a 2  b 2

Sol. Answer (2) asin2+ bcos2= m

…(1)

bsin2 + acos2 = n

…(2)

atan = btan(ii)

…(3)

Divide (1) by cos2 we, get atan2 + b = m sec2 2  tan  

mb am

…(4)

Divide (2) by cos2, we get Btan2 + a = nsec2 2  tan  

na bn

…(5)

From (3), (4), (5) ⎛ m  b⎞ ⎛ n  a⎞  b2 ⎜ a2 ⎜ ⎝ a  m ⎟⎠ ⎝ b  n ⎟⎠

 a2(mb – mn – b2 + bn) = b2(an – a2 – mn – am)  abm(a – b) + abn(a – b) = mn(a2 – b2)  abm + abn = mn(a + b) Divide both sides by abmn, we get

1 1 1 1    n m a b Transformation Formulae 38. The value of sin75° is equal to (1)

2 3 2

(2)

3 1 2 2

(3)

3 1 2 2

(4)

3 1 2 2

Sol. Answer (2) sin(A + B) = sinA cosB + cosA sinB

39. If tan  

(1)

 6

1 1 and tan   , then the value of  +  is 2 3

(2) 

(3) 0

(4)

 4


58



Sol. Answer (4) tan( + ) =

tan   tan  1 – tan  tan 

⎛ ⎞ ⎛ ⎞ 40. The value of cot ⎜   ⎟ cot ⎜   ⎟ is ⎝4 ⎠ ⎝4 ⎠ (1) –1

(2) 0

(3) 1

(4) Not defined

Sol. Answer (3) 41. The value of cos248° – sin212° is

5 1 8

(1)

(2)

5 1 8

5 1 5

(3)

5 1

(4)

2 2

Sol. Answer (1) cos2A – sin2B = cos(A + B) cos(A – B) 42. The value of sin(45° + ) – cos(45° –) is (1) 2cos 

(2) 2sin 

(3) 1

(4) 0

Sol. Answer (4) sin = cos(90° – ) 43. If cos(   ) 

(1)

4 5  and sin(  )  where ,  lie between 0 and , then the value of tan (2) is 5 13 4

25 16

(2)

56 33

(3)

19 12

(3)

3 2

(4)

20 7

Sol. Answer (2)

cos(  ) 

4 5

⇒

tan(  ) 

3 4

sin(  ) 

5 13

⇒

tan(  ) 

5 12

tan       tan(    ) tan 2  tan ⎡⎣      (   )⎤⎦  1  tan(   ) tan(  – )

44. The value of

(1) 1

1  tan2 15 is 1  tan2 15 (2)

3

(4) 2

Sol. Answer (3) 1  tan2 15 cos2 15  sin2 15 cos 30 cos0   1 1  tan2 15 cos2 15  sin2 15



3 2




59

45. If (1 + tan)(1 + tan) = 2, then  +  = (1) 30°

(2) 45°

(3) 60°

(4) 75°

Sol. Answer (2) 46. The value of cos12° + cos84° + cos156° + cos132° is (1)

1 2

(2) 1

(3)

1 2

(4)

(3)

1 2

(4) 2

1 8

Sol. Answer (3) cos12° + cos132° + cos84° + cos156° ⎛ 144 ⎞  ⎛ 120 ⎞  ⎛ 240 ⎞  ⎛ 72 ⎞  = 2cos ⎜⎝ ⎟ cos ⎜⎝ ⎟  2cos ⎜⎝ ⎟ cos ⎜⎝ ⎟⎠ 2 ⎠ 2 ⎠ 2 ⎠ 2

= 2cos72 

1  2cos120  cos36 2

= cos72° – cos36°

⎛ 108 ⎞  ⎛ 36 ⎞  = 2sin ⎜⎝ ⎟ sin ⎜⎝ ⎟⎠ 2 ⎠ 2 = – 2sin54°sin18° ⎛ 5  1⎞ ⎛ 5  1⎞ = 2 ⎜ ⎟⎜ ⎟ ⎝ 4 ⎠⎝ 4 ⎠

= 

1 2

47. The value of sin50° – sin70° + sin10° is equal to (1) 1

(2) 0

Sol. Answer (2) sin50° + sin10° – sin70° ⎛ 60 ⎞  ⎛ 40 ⎞  = 2 sin ⎜⎝ ⎟⎠ cos ⎜⎝ ⎟⎠  sin70 2 2

= (cos20°) – cos20° =0

48. The value of sin

(1) sin

  2 5  sin  sin  sin is 18 9 9 18

7 4  sin 18 9

(2) 1

(3) cos

 3  cos 6 7

(4) cos

   sin 9 9


60



Sol. Answer (1) Let  sin + sin2 + sin4 + sin5 = sin + sin5 + sin2+ sin4 = 2sin3cos2 + 2sin3cos = 2sin3 (cos2 + cos) = 2 sin = cos

3 ⎛ 2 ⎞  cos ⎟ ⎜⎝ cos 18 18 18 ⎠

   cos 9 18

⎛  ⎞ ⎛  ⎞ = sin ⎜⎝  ⎟⎠  sin ⎜⎝  ⎟⎠ 2 9 2 18

 sin

7 8  sin 18 18

= sin

7 4  sin 18 9


(1)

 13 sin is 10 10

1 2

(2)

1 2

(3)

1 4

(4) 1

 4

(4)

Sol. Answer (3)

3 ⎞ ⎛ sin18 sin ⎜   ⎟ ⎝ 10 ⎠ = – sin18°sin54° ⎛ 5  1⎞ ⎛ 5  1⎞ = ⎜ ⎟⎜ ⎟ ⎝ 4 ⎠⎝ 4 ⎠

= 

4 1  16 4

50. In a ABC, if sinA – cosB = cosC, then the measure of B is (1)

 2

(2)

 3

(3)

 6

Sol. Answer (1) sinA = cosC + cosB

sin A  2cos

B C B C .cos 2 2

⎛  A⎞ ⎛ B C⎞ sin A  2cos ⎜  ⎟ cos ⎜ ⎝ 2 2⎠ ⎝ 2 ⎟⎠ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


sin A  2sin


A ⎛ B C⎞ cos ⎜ ⎝ 2 ⎟⎠ 2

2sin

A A A ⎛ B C⎞ cos  2 sin cos ⎜ 0 ⎝ 2 ⎟⎠ 2 2 2

2sin

A⎛ A B C⎞ ⎜⎝ cos  cos ⎟ 0 2 2 2 ⎠

 cos

61

A ⎛ B C⎞  cos ⎜ 0 ⎝ 2 ⎟⎠ 2

A A ⎡ ⎤ ⎢∵ A  0 or 2  0 ⇒ sin 2  0⎥ ⎣ ⎦

A B C  2 2 A–B+C=0  A+C=B Also, A + B + C = 180° 2B = 180°  B = 90° or

 2

51. The value of cos10° cos20° cos40° is (1)

1 cot10 8

(2)

1 tan10 8

(3)

1 sec10 4

(4)

1 cosec10 4

Sol. Answer (1) Multiply and divide by

1 sin10 2

52. The value of the expression

⎛  ⎞ ⎛ 9 ⎞ ⎛ 3 ⎞ ⎛ 5 ⎞ 2cos ⎜ ⎟ cos ⎜ ⎟  cos ⎜ ⎟  cos ⎜ ⎟ is ⎝ 13 ⎠ ⎝ 13 ⎠ ⎝ 13 ⎠ ⎝ 13 ⎠ (1) 0

(2) –1

(3) 1

(4) 2

Sol. Answer (1) ⎛ ⎞ ⎛ 9 ⎞ ⎛ 3 ⎞ ⎛ 5 ⎞ 2cos ⎜ ⎟ cos ⎜ ⎟  cos ⎜ ⎟  cos ⎜ ⎟ ⎝ 13 ⎠ ⎝ 13 ⎠ ⎝ 13 ⎠ ⎝ 13 ⎠

⎛ 10 ⎞ ⎛ 8 ⎞ ⎛ 3 ⎞ ⎛ 5 ⎞ = cos ⎜⎝ ⎟⎠  cos ⎜⎝ ⎟⎠  cos ⎜⎝ ⎟⎠  cos ⎜⎝ ⎟⎠ 13 13 13 13 3 ⎞ 5 ⎞ ⎛ ⎛ ⎛ 3 ⎞ ⎛ 5 ⎞ = cos ⎜⎝   ⎟⎠  cos ⎜⎝   ⎟⎠  cos ⎜⎝ ⎟⎠  cos ⎜⎝ ⎟⎠ 13 13 13 13

⎛ 3 ⎞ ⎛ 5 ⎞ ⎛ 3 ⎞ ⎛ 5 ⎞ =  cos ⎜⎝ ⎟⎠  cos ⎜⎝ ⎟⎠  cos ⎜⎝ ⎟⎠  cos ⎜⎝ ⎟⎠ 13 13 13 13 =0 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

62



53. If sin + cos = 1, then the value of sin2 is equal to (1) 1

(2)

1 2

(3) 0

(4) –1

(3) 3

(4) 4

Sol. Answer (3) On squaring both the sides, we get (sin + cos)2 = 1  sin2 + cos2 + 2sincos = 1  sin2 = 0 1 1 and tan B  , then tan (2A + B) is equal to 2 3

54. If tan A  (1) 1

(2) 2

Sol. Answer (3) 2 tan A tan2 A   1  tan2 A

1 2 4 1 3 1 4 2

tan 2 A  tan B tan(2 A  B )  = 1  tan 2 A tan B

4 1  3 3 3 4 1 1  3 3

55. The value of tan3A – tan2A – tanA is equal to (1) tan3A tan2A tanA

(2) –tan3A tan2A tanA

(3) tanA tan2A – tan2Atan3A – tan3A tanA

(4) tan6A

Sol. Answer (1) 3A = 2A + A  tan3A = tan(2A + A)  4 and  lies in the III quadrant, then the value of cos is 2 5

56. If sin  

(1)

1

(2)

5

1 10

(3)

1 5

(4) 

1 10

Sol. Answer (3) 180° <  < 270°  90° <

 < 135° 2

This means 

cos

 lies in the second quadrant. 2

 will be negative. 2

If sin  

4 3 , then cos   5 5




63

Now, cos2A = 2cos2A – 1

cos   2cos2



 1 2

3   2cos2  1 5 2 2cos2 cos2

cos

 2  2 5

 1  2 5

 1  – [Since lies in the II quadrant] 2 5 2

57. The value of tan75° – cot75° = (2) 2  3

(1) 2 3

(3) 2  3

(4) 2 3

(3) tan 

(4) cot 

Sol. Answer (1) tan75  cot 75 

sin75 cos75  cos75 sin75

sin2 75  cos2 75 2cos(75  2)  = sin75 cos75 sin(75  2)

=

58. The value of

(1)

2cos150 2cos30  2 3 sin150 sin30

sin   sin 2 is 1  cos   cos 2

1 tan  2

(2)

1 cot  2

Sol. Answer (3) sin   sin2 sin (1  2cos ) sin   tan   = 2 cos  1  cos   cos 2 cos   2cos 

59. The value of 2sinA cos3A – 2sin3A cosA is (1) sin4A

(2)

1 sin 4 A 2

(3)

1 sin 4 A 4

(4) sin2A

Sol. Answer (2) 2sinAcosA(cos2A – sin2A) = sin2Acos2A =

1 sin 4 A 2


64


60. The value of tan

5   tan is 12 12

3

(1)


(2) 2 3

(3)

3 2

3 2

(4) 

Sol. Answer (2)

tan

5   tan 12 12

= tan75° – tan15° = cot15° – tan15°  2 3 2 3 2 3

⎛ ⎞ ⎟ is equal to ⎝ 2 ⎠

61. If sin + sin = a and cos – cos = b then, tan ⎜ (1) 

a b

(2) 

b a

(3)

a2  b2

(4)

a b

Sol. Answer (2) sin + sin = a ⎛   ⎞ ⎛   ⎞ ⎛   ⎞ ⎛   ⎞ ⇒ 2 sin ⎜ cos ⎜  2 sin ⎜ sin ⎜ b ⎝ 2 ⎟⎠ ⎝ 2 ⎟⎠ ⎝ 2 ⎟⎠ ⎝ 2 ⎟⎠ a ⎛   ⎞ ⇒ cot ⎜  ⎟ ⎝ 2 ⎠ b

b ⎛   ⎞ ⇒ tan ⎜  ⎝ 2 ⎟⎠ a 3 +1

62. If tan  =

3 1

, then the expression cos2 + (2 + 3)sin2 is

(1) 2  3

(2) –1

(3) 1



(4)  2  3



Sol. Answer (3) 3 1

tan  

3 1

 2  3  tan75

  = 75°





cos 2  2  3 sin 2





 sin2  3  2  3  2  sin50  2 

1 1 2




63. If 0 < 2 < 1 <

 3 , cos(1 + 2 ) = and 4 5

cos(1 – 2) =

4 , then sin21 equal to 5

(1) – 1

(2) 1

(3) 2

65

(4) – 2

Sol. Answer (2) cos  1  2  

3 4 , cos  1  2   5 5

sin(21) = sin[(1 + 2) + (1 – 2)] 

4 4 3 3 .  . 5 5 5 5



25 1 25

64. If sin  =

 1 1 ,cos  = , then    belongs to, where 0 < ,   2 2 3

⎛ ⎞ ⎝3 2⎠

⎛  2 ⎞ ⎟ ⎝2 3 ⎠

(1) ⎜ , ⎟

(2) ⎜ ,

⎛ 2 5 ⎞ , ⎟ ⎝ 3 6 ⎠

⎛ 5 ⎞ , ⎟ ⎝ 6 ⎠

(3) ⎜

(4) ⎜

(3) [–4, 4]

(4) [–10, 7]

Sol. Answer (2)

sin   



1 , 2

 , 6

cos   

1 3

 3

⎛  2 ⎞   ⎜ , ⎝ 2 3 ⎟⎠

65. The range of f() = 3cos2 – 8 3 cos·sin + 5sin2 – 7 is given by (1) [–7, 7]

(2) [–10, 4]

Sol. Answer (2) f() = 3cos2 – 8 3 cossin + 5sin2 – 7 = 3(cos2 + sin2) + 2sin2 – 4 3 sin2 – 7 = 3 + 1 – cos2 – 4 3 sin2 – 7 = – cos2 – 4 3 sin2 – 3 Maximum value =

1  48  3  7  3  4

Minimum value = – 7 – 3 = – 10 Rf = [– 10, 4] Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

66



⎛ ⎞ ⎛ ⎞   ⎟  2cos ⎜   ⎟ for real values of  is 4 4 ⎝ ⎠ ⎝ ⎠

66. The maximum value of 1  sin ⎜ (1) 3

(2) 5

(3) 4

(4) 2

Sol. Answer (3) ⎛ ⎞ ⎛ ⎞ 1  sin ⎜  ⎟  2cos ⎜  ⎟ ⎝4 ⎠ ⎝4 ⎠  1

1 2

1 ⎛1 ⎞ (sin   cos )  2 ⎜ cos   sin ⎟ ⎝2 ⎠ 2

⎛ 1 ⎞ ⎛ 1 ⎞  1  sin  ⎜  2 ⎟  cos ⎜  2⎟ ⎝ 2 ⎠ ⎝ 2 ⎠

 1

3 2

sin  

3 2

Maximum value =

cos  9 9  1 3 1 4 2 2

⎛⎞ ⎛⎞ ⎛  ⎞ ⎛  ⎞ ·cos ⎜ ⎟ ·cos ⎜ ⎟ ........cos ⎜ n ⎟ equals ⎟ ⎝4⎠ ⎝8⎠ ⎝ 16 ⎠ ⎝2 ⎠

67. The value of cos ⎜

⎛  ⎞ cosec ⎜ n ⎟ 2 ⎝2 ⎠ 1

(1)

n

(2)

1 n 1

2

⎛  ⎞ cosec ⎜ n 1 ⎟ ⎝2 ⎠

(3)

⎛  ⎞ cosec ⎜ n1 ⎟ 2 ⎝2 ⎠ 1

n

(4)

1 n 1

2

⎛  ⎞ cosec ⎜ n ⎟ ⎝2 ⎠

Sol. Answer (4)

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ cos ⎜ ⎟ .cos ⎜ ⎟ .cos ⎜ ⎟ ........cos ⎜ n ⎟ ⎝ 4⎠ ⎝ 8⎠ ⎝ 16 ⎠ ⎝2 ⎠ = cosA.cos2A.cos2nA ………. cos2n–2A., A 



1 n

2 sin A

 sin 2n 1 A



⎛ 2n A ⎞  sin ⎜ 2 ⎟ 2n sin  ⎝ ⎠



⎛ ⎞  sin ⎜ ⎟ ⎝ 2⎠ 2 sin 





 2n

1

1

n

1 ⎛ ⎞ 2n  sin ⎜ n ⎟ ⎝2 ⎠

1 2n

⎛ ⎞ cosec ⎜ n ⎟ ⎝2 ⎠




67

68. The expression tan + 2tan2 + 4tan4 + 8cot8 is equal to (if all terms are defined) (1) tan

(2) cot

(3)

1 tan 2 2

(4)

1 cot  2

Sol. Answer (2) tan + 2tan2 + 4tan4 + 8cot8 = cot – 2cot2 + 2cot2 – 4cot + 4cot – 8cot8 + 8cot8 = cot 69. If tanA + tanB + tanC = tanA·tanB·tanC, then (1) A, B, C must be angles of a triangle

(2) The sum of any two of A, B, C is equal to third

(3) A + B + C must be integral multiple of 

(4) A + B + C must be odd integral multiple of 

Sol. Answer (3) If A + B + C = n  tanA + tanB + tanC = tanA · tanB · tanC


 3 5 + sin + sin +........ to n terms is equal to n n n

(1) 1

(2) 0

(3)

n 2

(4)

n 1 2

(3)

n 1 2

(4)

n 1 2

Sol. Answer (2) sin

 3 5  sin  sin  ....... n n n

⎛ 2 ⎞ sin ⎜ n. ⎟ ⎝ 2n ⎠ 2 ⎞ ⎛   sin ⎜  (n  1)  ⎟  0 ⎝n 2n ⎠ ⎛ 2 ⎞ sin ⎜ ⎟ ⎝ 2n ⎠

n 1

71.

∑ cos2

r =1

(1)

r is equal to n

n 2

(2)

n 1 2

Sol. Answer (3) n 1

∑ cos2 r 1

 cos2

r n

 2 3 (n  1)  cos2  cos2  .........  cos2 n n n n

1⎡ 2 4 ⎤  ⎢(n  1)  cos  cos  ........⎥ upto n  1 terms 2⎣ n n ⎦ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

68



⎞ ⎛ sin ⎜ (n  1) ⎟ ⎝ ⎞ n 1 1 n⎠ ⎛ 2    (n  2) · ⎟ · cos ⎜ ⎝ n 2 2 n⎠ ⎛ ⎞ sin ⎜ ⎟ ⎝ n⎠



n 1 1 n   1 2 2 2

⎞ ⎞ ⎛ ⎛ 72. If tan  + tan ⎜  + ⎟ + tan ⎜   ⎟ = K tan3 , then K is equal to 3 3 ⎝ ⎠ ⎝ ⎠ (1) 1

(2) 3

(3)

1 3

(4) 2

1 64

(4)

Sol. Answer (2)

⎞ ⎞ ⎛ ⎛ tan   tan ⎜   ⎟  tan ⎜   ⎟  K tan3 ⎝ ⎝ 3⎠ 3⎠ K=3 73. The value of sin

 3 5 13 ·sin ·sin ........sin is equal to 14 14 14 14

(1) 1

(2)

1 16

(3)

1 32

Sol. Answer (3) sin

 3 5 7 9 11 13 .sin .sin .sin .sin .sin .sin 14 14 14 14 14 14 14

 sin2

 3 5 .sin2 .sin2 14 14 14

 3 5 ⎞ ⎛  ⎜ sin .sin .sin ⎟ ⎝ 14 14 14 ⎠

2

 2 3 ⎞ ⎛  ⎜ cos .cos .cos ⎟ ⎝ 7 7 7⎠

⎛ 1⎞ ⎜ ⎟ ⎝ 8⎠

2



2

1 64

74. In a triangle ABC, tanA + tanB + tanC = 6 and tanA·tanB = 2, then value of tanA, tanB and tanC are (1) 3, 1, 2

(2) 1, 2, 4

(3) 1, 2, 3

(4) 2, 2, 2

Sol. Answer (3) tanA + tanB + tanC = 6  tanA tanB tanC = 6

⇒

tan C 

6 3 2

tanA = 1, tanB = 2, tanC = 3 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



75. Let a = cosA + cosB – cos(A + B) and b = 4 sin

(1) 1

69

A B ⎛A+B⎞ ·sin ·cos ⎜ ⎟ , then a – b is equal to 2 2 ⎝ 2 ⎠

(2) Zero

1 2

(3) –1

(4)

(3) 2

(4) –2

Sol. Answer (1) a = cosA + cosB – cos(A + B)

 2cos

AB AB AB cos  2cos2 1 2 2 2

 1  2cos

AB⎛ AB A  B⎞  cos ⎜⎝ cos ⎟ 2 2 2 ⎠

 1  2cos

AB A B  2sin sin 2 2 2

 1  4cos

AB A B  sin sin 2 2 2

=1+b  a–b=1 76. The value of tan615° – 15tan415° + 15tan215° – 3 is (1) –1

(2) 1

Sol. Answer (4) Let  = 15°  3 = 45° tan 3 = 1 3 tan  – tan3  = 1 – 3 tan2  Squaring both side 9tan2  + tan6  – 6tan4  = 1 + 9tan4  – 6tan2   tan6  – 15tan4  + 15tan2  – 3 = –2 77. If A, B, C are in A.P., then correct relation is (1) cot A =

sin B  sin C cos C  sin B

(2) cot B =

sin A  sin C cos C  cos A

(3) tan A =

sin A  sin C cos C  cos B

(4) tan B =

sin B  sin C cos C  cos B

Sol. Answer (2) A, B, C  A.P.  2B = A + C  B

AC 2


70



⎛ A C⎞ cot B  cot ⎜ ⎝ 2 ⎟⎠ ⎛ A  C⎞ cos ⎜ ⎝ 2 ⎟⎠  ⎛ A  C⎞ sin ⎜ ⎝ 2 ⎟⎠

⎛ A C⎞ ⎛ A C ⎞ cos ⎜ · sin ⎜ ⎝ 2 ⎟⎠ ⎝ 2 ⎟⎠  ⎛ A C⎞ ⎛ A C⎞ sin ⎜ · sin ⎜ ⎝ 2 ⎟⎠ ⎝ 2 ⎟⎠

78.

cos



sin A  sinC cosC  cos A

 3 5  + cos + cos  K cot , then K is equal to 14 14 14 14

(1) 1

(2)

1 2

(3) 2

(4) – 2

Sol. Answer (2)

cos

 3 5   cos  cos  K cot 14 14 14 14

 cos

 4    2cos  cos  K cot 14 14 14 14

 4  2  cos cos  14 14

 2cos

 sin

 14  cos   14 sin 14

K .cos

4    sin  K  sin 14 14 14

5 3   sin  K  sin 14 14 14

 K  sin

 2 sin

5 3   sin  sin 14 14 14

3 2 3  sin cos 14 14 14

 2 sin

3   2cos2 14 14

 4 sin

 3  cos2 14 14

⇒

K

1 2




71

79. The minimum value of 8(cos2 + cos) is equal to (1) –17

(2) –9

(3) 3

(4) –8

Sol. Answer (2) 8(cos2 + cos) = 8(2cos2 – 1 + cos) = 8(2cos2 + cos – 1) 1 1⎞ ⎛  16 ⎜ cos2   cos   ⎟ ⎝ 2 2⎠ ⎛⎛  16 ⎜ ⎜ cos   ⎝⎝

2 1⎞ 6⎞ ⎟⎠  ⎟ 4 16 ⎠

Minimum value = 16  

9 –9 16

80. If 4n = , then cot·cot2·cot3 ..... cot(2n – 1) n Z is equal to (1) 1

(2) –1

(3) 2

(4) Zero

Sol. Answer (1) 4n =   2n =

 2

cot.cot2.cot3 …….cot(2n – 1)

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ = cot.cot2.cot3 …….. cot ⎜⎝  3 ⎟⎠ .cot ⎜⎝  ⎟⎠ .cot ⎜⎝  ⎟⎠ 2 2 2 =1

1 81. The value of tan7  is equal to 2

(1)

3 2

(2)

2 +1

3 2 2 1

(3)

3+ 2 2 +1

(4)

3+ 2 2 1

Sol. Answer (1)



tan  

Put  

1  cos 2 sin2

15 2

15 1  cos15 tan   2 sin15

1

3 1 2 2 3 1 2 2


72




2 2  3 1 3 1




3 1 3 1



2 6  3  3  2 2  3 1 3 1



2 6 42 3 2 2 2

 6 4 3 2 

3  2  2 1



 6 4 3 2

3 2





2 1

3 2 2 1

82. If tan   a , where a is a rational number which is not a perfect square, then which of the following is a rational number? (1) sin2

(2) tan2

(3) cos2

(4) cosec2

Sol. Answer (3)

1

+

a

tan   a 

a

a  sin 2   1 a 1 a

a

 1

2 a sin2  1 a cos 2  2cos2   1  2 



1 1 1 a

2  1 a 1 a  1 a a 1

83. The value of 2tan18° + 3sec18° – 4cos18° is (1) Zero

(2)

5

(3)  5

(4)

3

Sol. Answer (1) 2tan18° + 3sec18° – 4cos18°



2sin18  3  4cos2 18 cos18



2sin18  3  4  4 sin2 18 cos18



4 sin2 18  2sin18  1 cos18



 4







5 1 16

2

 2.

cos18


73

5 1 1 4

62 5 2 5 24 0 4cos18

84. If tan2 – 2tan2 = 1, then which of the following is correct? (1) 2cos2 – cos2 = 1

(2) cos2 – 2cos2 = 1

(3) 2cos2 – cos2 = – 1

(4) cos2 – 2cos2 = –1

Sol. Answer (3) tan2 – 2tan2 = 1  tan2 = 1 + 2tan2  1 + tan2 = 2 + 2tan2  sec2 = 2sec2  cos2 = 2cos2 

2cos2 = 4sin2

 1 + cos2 = 2(1 + cos2)  2cos2 – cos2 = – 1 85. The value of tan10°·tan50°·tan70° is (1)

3

(2)

1 3

(3) 1

(4) –1

Sol. Answer (2) tan30° =

1 3

86. The expression 2sin2° + 4sin4° + 6sin6° + ........ 180sin180° equals (1) cot1°

(2) 90cot1°

(3) sin1°

(4) 90cos1°

Sol. Answer (2) 2sin2° + 4sin4° + 6sin6° +………….+ 178sin178° + 180sin180° = (2 + 178)sin2° + (4 + 176)sin4° + (6 + 174)sin6° + …….+ sin88° = 180°(sin2° + sin4° + sin6° + ……) + 90sin90°

2⎞ ⎛ sin ⎜ 44  ⎟ ⎝ 2⎞ 2⎠ ⎛  180   sin ⎜ 2  (44  1). ⎟  90 ⎝ ⎠ 2 2 ⎛ ⎞ sin ⎜ ⎟ ⎝ 2⎠

 180 

sin 44  sin 45  90 sin1


74




180 2




sin(45  1)  90 sin1

1 ⎡ 1 ⎤ cos1  sin1 ⎥ 180 ⎢ 2 2 ⎢ ⎥  90  sin1 2 ⎢ ⎥ ⎢⎣ ⎥⎦ 

180 ⎡ 1 1 ⎤ cot1  ⎢ ⎥  90 2 ⎣ 2 2⎦

 90 cot1 

180  90 2

= 90cot1° 87. If cos2x + 2cosx = 1, then sin2x(2 – cos2x) is equal to (1) 1

(2) –1

(3)  5

(4)

5

Sol. Answer (1) cos2x + 2cosx = 1 2cosx = 1 – cos2x 2cosx = 2sin2x cosx = sin2x  sin2x(2 – cos2x) = sin2x(1 + sin2x) = sin2x + sin4x = sin2x + cos2x =1 88. The value of tan (1) tan2n

 (1 + sec)(1 + sec2) (1 + sec22) ........ (1 + sec2n) is 2 (2) tan2n – 1

(3) tan2n + 1

(4) tan2n – 2

Sol. Answer (1) tan

 1  sec  (1  sec 2)(1  sec 4)........(1  sec 2n ) 2

 tan

 1  tan  1  tan2 1  tan 4    2 cos  cos 2 cos 4

  2cos2 2 2 2  2  2cos   2cos 2   cos  cos 2 cos 4 cos 2 sin

  cos .cos 2  23  sin .cos . 2 2 cos 4 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


 22  sin .cos .

2


75

cos 2 cos 4

sin2 cos 2 cos 4

= tan4 = tan22

  tan (1  sec )(1  sec 2).........(1  sec 22n ) 2 = tan2n 44

89. The value of [100(x – 1)] is where [x] is the greatest integer less than or equal to x and x =

∑ cos n°

n 1 44

∑ sin n°

n 1

(1) 140

(2) 141

(3) 142

(4) 144

Sol. Answer (2)

 A

44

∑ cos n

n 1

= cos1° + cos2° + cos3° + …….. + cos44° B = sin1° + sin2° + sin3° + ……. + sin44° x



A cos1  cos 2  cos3  .......  cos 44  B sin1  sin 2  sin3  ..........  sin 44

sin89  sin88  sin87  .......  sin 46 sin1  sin 2  sin3  ..........  sin 44

1⎞ ⎛ sin ⎜ 44. ⎟ ⎝ 1⎞ 2⎠ ⎛  cos ⎜ 1  (44  1)  ⎟ ⎝ 2⎠ ⎛ 1⎞ sin ⎜ ⎟ ⎝ 2⎠  1⎞ ⎛ sin ⎜ 44. ⎟ ⎝ 1⎞ 2⎠ ⎛  sin ⎜ 1  (44  1) ⎟ 1 ⎝ 2⎠ sin 2 1⎞ ⎟ 1 2⎠   cot 22  1⎞ 2 ⎛ sin ⎜ 22  ⎟ ⎝ 2⎠ ⎛ cos ⎜ 22  ⎝

 2 1

 ⎡100 ⎣





2  1  1 ⎤  100  1.41  141 ⎦


76



 and  +  = , then tan equals 2

90. If    

(1) 2(tan + tan)

(2) tan + 2tan

(3) tan – tan

(4) tan – tan

Sol. Answer (2)

 



 ,    2

  2

=–

 tan = cot

tan  

tan   tan  1  tan  tan 

 tan + tan tan tan = tan – tan  tan = tan – 2tan  tan = tan + 2tan 91. If f(, ) = cos2 + sin2·cos2, then which of the following is incorrect?

⎛  2 ⎞ ⎛ 2  ⎞ (1) f ⎜ , ⎟  f ⎜ 5 , 5⎟ 5 5 ⎝ ⎠ ⎝ ⎠

⎛  ⎞ ⎛  ⎞ (2) f ⎜ , ⎟ = f ⎜ , ⎟ 12 3 ⎝ ⎠ ⎝ 3 12 ⎠

⎛ ⎞ ⎛ ⎞ (3) 3f ⎜ , ⎟  f ⎜ , ⎟ 5 3 ⎝ ⎠ ⎝3 5⎠

⎛  ⎞ ⎛  ⎞ (4) f ⎜ , ⎟  3f ⎜ 18 , 4 ⎟ 4 18 ⎝ ⎠ ⎝ ⎠

Sol. Answer (1) f(, ) = cos2 + sin2.cos2 = 1 – sin2 + sin2.cos2 = 1 – (1 – cos2)sin2 = 1 – (1 – cos2)sin2 = 1 – 2sin2 sin2   ⎛  ⎞ f ⎜ , ⎟  1  2sin2 sin2 ⎝ 12 3 ⎠ 12 3

  1 2   1

 1



3 1

2



8



3 42 3

3 4



16



3 2 3 8

  863 8

3



23 3 8




77

92. Let f(x) = cos10x + cos8x + 3cos4x + 3cos2x and g(x) = 8cosx·cos33x, then for all x we have (1) f(x) = g(x)

(2) 2f(x) = 3g(x)

(3) f(x) = 2g(x)

(4) 2f(x) = g(x)

Sol. Answer (1) f(x) = cos10x + cos8x + 3cos4x + 3cos2x g(x) = 8cosx.cos33x = 8cosx(4cos3x – 3cosx)3 = 8cosx ((4cos3x)3 – 3.(4cos3x)2.3cosx + 3.(4cos3x)(3cosx)2 – (3cosx)3) = 8cosx(64cos9x – 144cos7x + 108cos5x – 27cos3x) = 8cos4x(64cos6x – 144cos4x + 108cos2x – 27)  f(x) = cos10x + cos8x + 3cos4x + 3cos2x n

93.

⎛

⎞

1

∑ ⎜⎝ cos  + cos(2r + 1) ⎟⎠ , n N is equal to

r =1

sin(n + 1) sin ·cos n

(1)

(2)

sin n sin2 ·cos(n + 1)

(3)

tan(n + 1) sin n

(4)

sin(n  1) sin ·cos n

Sol. Answer (2) n

⎛

1

⎞

∑ ⎜⎝ cos   cos(2r  1) ⎟⎠ r 1



1 1 1   .........  cos   cos3 cos   cos5 cos   cos(2n  1)



1 1 1   .........  2cos 2  cos  2cos3  cos 2 2cos(n  1).cos n



1 ⎡ sin(2  ) sin(3  2) ⎤   ........⎥ ⎢ 2sin  ⎣ cos 2.cos  cos3 cos 2 ⎦



1 ⎡ sin 2 cos   cos 2 sin  sin3 cos 2  cos3 sin 2 ⎤   ........⎥ 2 sin  ⎢⎣ cos 2.cos  cos3  cos 2 ⎦



1 [tan2 – tan + tan3 – tan2 + ……… tan(n + 1) – tann] 2sin 



1 [tan(n + 1) – tan] 2sin 



1 sin(n  1  1)  2sin  cos(n  1) cos 



sin n cos(n  1).sin2


78



94. The minimum value of 27cos 3x.81sin 3x is (1) 1

(2)

1 81

(3)

1 243

(4)

1 27

Sol. Answer (3) Let y = 27cos 3x.81sin 3x = 33cos 3x.34sin 3x = 33cos 3x + 4 sin 3x Then minimum value of y is 3–5 =

1 243 o

35

95. Given that

⎛m⎞ ∑ sin5k   tan ⎜⎝ n ⎟⎠ , where m and n are relatively prime positive integers that satisfy k 1

o

⎛m⎞ ⎜ n ⎟  90 , then m + n is equal to ⎝ ⎠ (1) 173

(2) 175

(3) 177

(4) 179

Sol. Answer (3) 35

m

∑ sin5k  tan n

k 1

 sin5 + sin10 + ….. + sin5.35 = tan

m n

5⎞ ⎛ sin ⎜ 35  ⎟ ⎝ 5⎞ m 2⎠ ⎛  sin ⎜ 5  (35  1)  ⎟  tan  ⎝ 2⎠ n ⎛ 5⎞ sin ⎜ ⎟ ⎝ 2⎠ ⎛ 175 ⎞ sin ⎜ ⎝ 2 ⎟⎠ m ⎛ 175  5 ⎞  sin ⎜  tan  ⎟ ⎝ 2 ⎠ n ⎛ 5 ⎞ sin ⎜ ⎟ ⎝ 2⎠ ⎛ 175 ⎞ sin ⎜ ⎝ 2 ⎟⎠ m  tan  n ⎛ 5 ⎞ sin ⎜ ⎟ ⎝ 2⎠ ⎛ 175 ⎞ ⎛ m⎞ sin ⎜ sin ⎜ ⎟ ⎝ 2 ⎟⎠ ⎝ n⎠   ⎛ 175 ⎞ ⎛ m⎞ cos ⎜ cos ⎜ ⎟ ⎝ 2 ⎟⎠ ⎝ n⎠ 

m 175  n 2

 m + n = 175 + 2 = 177 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


96. For 


79

  sin  + sin2 << , lies in the interval 2 2 1 + cos  + cos 2

(1) (–, )

(2) (–2, 2)

(3) (0, )

(4) (–1, 1)

Sol. Answer (1)

sin   sin 2 1  cos   cos 2



sin (1  2cos ) cos (1  2cos )

= tan ; tan  (–, )

A B C⎞ C ⎛ 97. If A + B + C =  and sin ⎜ A + ⎟ = K sin , then tan · tan 2 2 2 2 ⎝ ⎠ (1)

K 1 K +1

(2)

K 1 K –1

(3)

K K +1

(4)

K 1 K

Sol. Answer (1) A+B+C= C⎞ C ⎛ sin ⎜ A  ⎟  K sin ⎝ 2⎠ 2

C⎞ ⎛ sin ⎜ A  ⎟ ⎝ 2⎠ K   C 1 sin 2 C⎞ C ⎛ sin ⎜ A  ⎟  sin ⎝ ⎠ 2 2 K 1   C⎞ C K 1 ⎛ sin ⎜ A  ⎟  sin ⎝ 2⎠ 2

A ⎛ A C⎞ 2 sin ⎜ cos ⎟ ⎝ 2 ⎠ 2 K 1   A C A  K 1 ⎛ ⎞ 2cos ⎜ sin ⎟ ⎝ 2 ⎠ 2 B A cos 2. 2  K 1  B A K 1 sin sin 2 2 cos

 sin

A B K 1 .sin  2 2 K 1


80



98. If A + B + C =  and cosA = cosB·cosC, then tanB·tanC is equal to (1) 1

(2)

1 2

(3) 2

(4) 3

(3) 45°

(4) 60°

Sol. Answer (3) A+B+C= cosA = cosB.cosC  cos(B + C) = – cosB cosC  cosB cosC – sinB sinC = –cosB cosC  sinB sinC = 2cosB cosC  tanB tanC = 2 Trigonometric Equations 99. If 2cos2 – 2sin2 = 1, then  is equal to (1) 15°

(2) 30°

Sol. Answer (2) 2(cos2 – sin2) = 1 2cos2 = 1 cos 2 

1  cos 60 2

2 = 60°  = 30°

1 , then the value of  is 4

100. If sin2   (1) n  

 6

(2) n  

 4

(3) n  

 3

(4) n

Sol. Answer (1) 101. The solution of the equation cos2 + sin + 1 = 0 lies in the interval

⎛   ⎞ (1) ⎜ , ⎟ ⎝ 4 4⎠

⎛ 3 5 ⎞ (3) ⎜ , ⎟ ⎝ 4 4 ⎠

⎛  3 ⎞ (2) ⎜ , ⎟ ⎝4 4 ⎠

⎛ 5 7 ⎞ (4) ⎜ , ⎟ ⎝ 4 4 ⎠

Sol. Answer (4) cos2 + sin + 1 = 0 1 – sin2 + sin + 1 = 0 sin2 – sin – 2 = 0 (sin – 2) (sin + 1) = 0  sin = – 1     

∵ sin   2

⇒ sin  – 2  0

 3 6   2 2 4

 6  ⎛⎜ 5 , 7 ⎞⎟ 4 ⎝ 4 4⎠ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



81

102. The general solution of tan3x = 1 is  4

(1) n  

(2)

n   3 12

(3) n

(4) n  

 4

Sol. Answer (2) 103. If sin   3 cos , –  < < 0, then the value of  is (1)

5 6

(2)

4 6

(3)

2 3

(4)

5 6

Sol. Answer (2) sin   3 cos 

 tan   3

⇒   n 

 3

Taking n = – 1, we get    

 2 4   3 3 6

104. The total number of solutions of the equation tanx + secx = 2 which lie in the interval [0, 2] is (1) 0

(2) 1

(3) 2

(4) 3

Sol. Answer (2) secx = 2 – tanx sec2x = 4 + tan2x – 4tanx 1 + tan2x = 4 + tan2x – 4tanx 4tanx = 3 3 4

tan x 

Since tanx is positive in I and III quadrant only, thus tan x  105. If (1)

3 for 2 values of x. 4

3 sec   2  0 , then principal value of  may be 5 6

(2) 

5 6

(3)

7 6

(4) 

7 6

Sol. Answer (1)

sec   

2 3

 cos   

  

3 2

5 (Principal) 6


82



106. The number of solutions of the equation2tan2 – 7sec – 2 = 0 in the interval [0, 6] is (1) 8

(2) 6

(3) 4

(4) Zero

Sol. Answer (2) 2(1 + sec2) – 7sec – 2 = 0  2sec2 – 7sec = 0  sec  

7 2

In every interval of 2, 2 solutions exist. In 6 interval, 6 solutions will exist. 107. The general solution of the equation tanx + tan2x + tanx· tan2x = 1 is (1) x  n 

 5 , n  , n  I 12 12

(2) x 

  (3) x  n  , n  , n I 4 12

n   , n I 3 12

 5 (4) x  n  , n  , n I 4 12

Sol. Answer (2)

tan x  tan2 x 1 1 tan x tan2 x  tan3x = 1  3 x  n 

 x 

 4

n   3 12

108. If angles A and B satisfy (1 + tanA)(1 + tanB) = 2, then the value of A + B is (1) 

11 4

(2)

15 4

(3)

23 4

(4) 

17 4

Sol. Answer (1) 1 + tanA + tanB + tanA tanB = 2 

tan A  tan B 1 1 tan A tan B

 tan(A + B) = 1 

A  B  n 

 4

For n = –3, A  B 

11 4




83

109. The number of solution of the equation tan3x – tan2x – tan3x · tan2x = 1 in [0, 2] is (1) 1

(2) Zero

(3) 3

(4) 2

Sol. Answer (2) tan3x – tan2x = 1 + tan3x tan2x 

tan3 x  tan2 x 1 1 tan3 x tan2 x

 tanx = 1 

x  n 

 4

 5 x  , , ........ 4 4

But for each of values tan2x, tan3x will not be defined. Therefore no solution exists. Trick : Always check the values by back substitution in case of tanx. 110. If x  (0, 1), then greatest root of the equation sin2x  2cos x is

(1)

1 4

(2)

1 2

(3)

3 4

(4)

1 3

Sol. Answer (3) x  (0, 1) sin2x  2 cos x

 2sin x cos x  2 cos x  2 cos x ⎡⎣ 2 sin x  1⎤⎦  0  cosx = 0

x  (2n  1)

x

 2

(2n  1) 2

1 5 x  , , ........ 2 2

or

sin x 

1 2

x  n  ( 1)n

x n

 4

( 1)n 4

3 1 9 x  , , ........ 4 4 4

Greatest root in (0, 1) interval is

3 . 4


84



111. The number of solutions of equation 3cos2 + 5cos = 1 in [0, 2] is (1) 8

(2) 6

(3) 4

(4) 2

Sol. Answer (4) 3cos2 + 5cos = 1  3(2cos2 – 1) + 5cos = 1  6cos2 + 5cos – 4 = 0  6cos2 + 8cos – 3cos – 4 = 0  2cos(3cos + 4) – 1(3cos + 4) = 0  (2cos – 1) (3cos + 4) = 0  cos  

1 2

or

In [0, 2], two

cos  

4 3

Not possible solutions exist

112. The general solution of the equations tan = –1 and cos  

(1) n 

7 4

n (2) n  ( 1)

7 4

1 is (n  I) 2

(3) 2n 

7 4

(4) 2n 

5 4

Sol. Answer (3)

tan   1, cos  

1 2

In [0, 2] interval, tan = –1  

3 7 , 4 4

In [0, 2] interval, cos    

1 2

 7 , 4 4

Common  

7 4

 General solution : 2n 

7 4

113. The general solution of sec   tan   3 is   (1)   2n  , 2n  , n I 6 2

(3) 2n  

 ; n I 2

 (2)   2n  , n  I 6

(4) 2n  

 ; n  2k  1, n, k  I 2




85

Sol. Answer (2)

sec   tan   3 We know that sec2 – tan2 = 1 (sec – tan) (sec + tan) = 1

sec   tan  

1 3

sec   tan   3

2sec   3 

sec  

1 3



4 3

2 3

 cos  

3 2

   2n 

 6

114. The number of the solutions of the equation 3sinx + 4cosx – x2 – 16 = 0 is (1) 3

(2) 2

(3) 1

(4) 0

Sol. Answer (4) 3sinx + 4cosx = x2 + 16 Range of LHS = [–5, 5] Range of RHS = [16, )  No solution exists. 115. The solution set of sin4x – tan8x = 1 is given by  (1) x  2n  , n I 8

(2) x  n 

 , n I 12

(3) x  2n 

 , n I 24

(4) None of these

Sol. Answer (4) sin4x – tan8x = 1  sin4x = 1 + tan8x Range of LHS : [0, 1] Range of RHS : [1, ] Only solution exists when LHS and RHS are both equal to 1. sin4x = 1, 1 + tan8x = 1 sin2x = 1, tan8x = 0 

x  n 

 but at these value tan8x 0. 2


86



116. If   (0, 2) and 2sin2 – 5sin + 2 > 0, then the range of  is

⎛  ⎞ ⎛ 5 ⎞ (1) ⎜ 0, ⎟  ⎜ , 2 ⎟ ⎝ 6⎠ ⎝ 6 ⎠

⎛ ⎞ (2) ⎜ 0, ⎟  ( , 2) ⎝ 6⎠

⎛ ⎞ (3) ⎜ 0, ⎟  ( , 2) ⎝ 6⎠

(4) 

Sol. Answer (1)   (0, 2) 2sin2 – 5sin + 2 > 0  2sin2 – 4sin – sin + 2 > 0  2sin(sin – 2) – 1(sin – 2) > 0

(2sin   1) (sin   2)  0  ve

1 2 O

 2sin – 1 < 0  sin  

1 2

 6

5  6

2

⎛  ⎞ ⎛ 5 ⎞  ⎜ 0, ⎟  ⎜ , 2⎟ ⎝ 6⎠ ⎝ 6 ⎠

117. The general solution of the equation tan2 tan3 = 1 is (1)  = (4n + 1)

 , n  4k – 2, k  I 10

(2)  = (2n + 3)

 , n  5k + 2, k  I 10

(3)  = (4n + 3)

 , n  5k + 2, k  I 10

(4)  = (2n + 1)

 , n  5k – 3, k  I 10

Sol. Answer (4)

118. If m, n  N(n > m), then number of solutions of the equation n| sinx | = m| sinx | in [ 0, 2] is (1) m

(2) n

(3) mn

(4) 3

(3) A = 2n+ B, n  I

(4) A = n– B, n  I

Sol. Answer (4) n|sinx| = m|sinx|  (n – m) |sinx| = 0 Since n > m  |sinx| = 0  x = n In [0, 2], x = 0, , 2 119. If sinA = sinB and cosA = cosB, then (1) A = B + n, n  I

(2) A = B – n, n  I




87

Sol. Answer (3) sinA = sinB, cosA = cosB A = m + (–1)n. B, A = 2k ± B Clearly common solution is A = 2n + B, n  I 120. The number of values of x  [–2, 2] satisfying tanx + cotx = 2cosecx is (1) 2

(2) 4

(3) 6

(4) 8

Sol. Answer (2) tan x  cot x  2cosec x

1/2

1 2  sin x cos x sin x

 cos x 

1 2

(∵ sin x  0)

 Number of solutions in [–2, 2] equals 4. 121. The solution set of the equation tanax = tanbx where (a  b) constitutes (1) An A.P. with common difference

  or ab ab

(2) An H.P. with common difference

 |a  b|

(3) An A.P. with common difference

 ab

(4) An A.P. with common difference

  or ab b a

Sol. Answer (4) tan ax = tan bx  ax = n + bx 

x

n , n Z ab

 A.P. with common difference 

  or (a  b ) (b  a )

122. The general solution of the equation

(1) n

(2)

n 3

tan x tan2 x   2  0 is tan2 x tan x  (3) (2n  1) , n  I 3

 (4) (3n  1) , n I 3


88



Sol. Answer (4)

tan x tan2 x  20 tan2 x tan x

tan x(1 tan2 x ) 2tan x  20 2tan x (1 tan2 x )tan x If tanx  0,

1 tan2 x 2  20 2 1 tan2 x

2 2  (1 tan x )  4  2  0 2(1 tan2 x )







1 tan4 x  2tan2 x  4  4  4tan2 x 2(1 tan2 x ) tan4 x  6tan2 x  9 2(1 tan2 x ) (tan2 x  3)2 2(1 tan2 x )

 x  n 

0

0

 0 ⇒ tan2 x  3

 3

123. The solution of the equation cos2 – 2cos = 4sin – sin2 where   [0, ] is (1)  – , tan  

1 2

(2)  – , cot   

1 2

(3)  – , tan = 2

(4)  + , cot = 2

Sol. Answer (1) cos2 – 2cos = 4sin – sin2  cos2 – 2cos = 4sin – 2sin cos  cos(cos – 2) + 2sin(cos – 2) = 0  (cos + 2sin) (cos – 2) = 0  cos = 2 Not possible

or

cos = –2sin

tan   

1 2

= n + 

1 ⎛ 1⎞  n  tan ⎜⎝ ⎟⎠ 2

1 = n  tan

1 2

1 For n = 1,     tan

1 2




89

124. The general solution of sin2  sec   3 tan   0 is  (1)   n  ( 1)n 1 ,   n, n I 3

(2)  = n, n I

 (3)   n  ( 1)n 1 , n I 3

(4)  =

n , n I 2

Sol. Answer (2)

sin2  sec   3 tan   0 sin2   3 sin  0 cos   sin (sin   3)  0, cos   0 =n,nI

125. If sin x  cos x  y 

(1) x 

1 , x(0, ) , then y

 ,y=1 4

(2) y = 0

(3) y = 2

(4) x 

3 4

Sol. Answer (1)

sin x  cos x  y 

1 y

x  (0, )

Comparing range on both sides, LHS = ⎡⎣  2, 2 ⎤⎦ RHS = ⎡⎣ 2, 



Only solution exists sin x  cos x  2 ⎞ ⎛ sin ⎜ x  ⎟  1 ⎝ 4⎠

 x

   n  ( 1)n 4 2

 x

 4

Also,

y

1  2 ⇒ y 1 y


90



126. The number of solutions of the equation cos(  x  4) cos(  x )  1 is (1) Zero

(2) 1

(3) 2

(4) Infinite

(3) 6

(4) 12

Sol. Answer (2) cos  x  4 cos  x  1

Only possible when cos  x  4  1

 x  4  2n  x 4 

cos  x  1

and

 2

 x  2m 

(2n  1) 2

x

1 5 7 9 x4  , , , 2 2 2 2

 2

(2m  1) 2

1 5 7 x  , , ........ 2 2 2

 No solution exists. 6

127. The number of solutions of

∑ cos(rx )  6

in (0, 2] is

r 1

(1) Zero

(2) 1

Sol. Answer (2) cosx + cos2x + …….. + cos6x = 6  cosx = 1, cos2x = 1, cos3x = 1, …….., cos6x = 1

cos x  1

cos2 x  1

cos3 x  1

x  0, 2

x  0,  , 2

x

.....................

2 4  , , 2 3 3

cos6 x  1 ⇒ x  2

 x = 2 satisfies all conditions. 128. The number of values of x for which sin2x + cos4x = 2 is (1) Zero

(2) 1

(3) 2

(4) Infinite

Sol. Answer (1) sin2x + cos4x = 2  sin2x = 1 , n  2 x  n  ( 1)

cos4x = 1  2

n   ( 1)n 2 4



x



 5 x  , ........ 4 4

4x = 2n x

n 2

 3 x  ,  , ........ 2 2

 No solution exists. Except x = 4 one solution. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



91

129. If the equation cosx + 3cos(2Kx) = 4 has exactly one solution, then (1) K is a rational number of the form

P , P  –1 P 1

(2) K is irrational number whose rational approximation does not exceed 2 (3) K is irrational number (4) K is a rational number of the form

P ,P1 P 1

Sol. Answer (3) cosx + 3cos 2kx = 4  cosx = 1,

cos2kx = 1

 x = 2n

2kx = 2m

x

Comparing 2n 

m k

m m ⇒ k k 2n

For infinite m, n  integer, k will be rational but equation has only 1 solution, which is only possible when k  irrational. 130. The solution of the inequality log1/2sinx > log1/2cosx is

⎛ ⎞ (1) x  ⎜ 0, ⎟ ⎝ 2⎠

⎛ ⎞ (2) x  ⎜ 0, ⎟ ⎝ 8⎠

⎛ ⎞ (3) x  ⎜ 0, ⎟ ⎝ 4⎠

⎡ ⎤ (4) x ⎢0 , ⎥ ⎣ 4⎦

Sol. Answer (3) log1/2 sinx > log1/2 cosx  sinx < cosx

O  4

 2



3 2 2

⎛ ⎞ From graph sinx < cosx in ⎜⎝ 0, ⎟⎠ 4

131. The values of k for which the equationsin4x + cos4x + sin2x + k = 0 possesses solution is

⎡ 3 ⎤ (1) k  ⎢ ,1⎥ ⎣ 2 ⎦

⎡ 1⎤ (2) k  ⎢0, ⎥ ⎣ 2⎦

⎡ 3 1⎤ (3) k  ⎢ , ⎥ ⎣ 2 2⎦

⎡1 ⎤ (4) k  ⎢ ,1⎥ ⎣2 ⎦

Sol. Answer (3) sin4x + cos4x + sin2x + k = 0 1 – 2sin2x cos2x + 2sinx cosx + k = 0 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

92



1 2 sin 2 x  sin2 x  (k  1)  0 2

D0 ⎛ 1⎞ 1 4 ⎜ ⎟ (k  1)  0 ⎝ 2⎠

1 + 2k + 2  0

 2k + 3  0 

k

3 2

sin22x – 2sin2x – 2(k + 1) = 0 sin22x – 2sin2x – 2 = 2k (sin2x – 1)2 – 3 = 2k  0  2k + 3  1  –3  2k  –2 

3  k  1 2

132. The solution set of the equation 4sin4x + cos4x = 1 is (1) x  n 

 6

(3) x  (2n  1)

(2) x = n ±  where cos2    2

(4) x  2n 

3 and x = n, n  I 5

 3

Sol. Answer (2) 4sin4x + cos4x = 1  4sin4x + (1 – sin2x)2 = 1  4sin4x + 1 + sin4x – 2sin2x = 1  5sin4x – 2sin2x = 0  sin2x(5sin2x – 2) = 0 2 2  sin x  0, sin x 

2 5

2  x = n or x = n ± , where sin  

2 3 , cos2   , n I 5 5

133. The number of values of   [0, 2] satisfying r sin   3 and r  4sin   2( 3  1) is (1) 4

(2) 5

(3) 6

(4) 7




93

Sol. Answer (1) r sin   3

r  4sin   2( 3  1) r

4 3 2 3 2 r

 r 2  2( 3  1)r  4 3  0

r (r  2 3)  2(r  2 3)  0 r  2 or r  2 3 sin  

3 1 , sin   2 2

 4 solutions exist in [0, 2] 134. The arithmetic mean of the roots of the equation 4cos3x – 4cos2x – cos(315 + x) = 1 in the interval (0, 315) is equal to (1) 50

(2) 51

(3) 100

(4) 315

Sol. Answer (2) 4cos3x – 4cos2x + cosx = 1  4cos3x – 4cos2x + cosx – 1 = 0  (4cos2x + 1) (cosx – 1) = 0  cosx = 1  x = 2n, n I x = 0, 2, 4, …….. 100

AM =

2  4  ........  100 50  [102]  51 50 2  50

135. The equation 2sin2x = 8 – k(2 – sinx) possesses a solution if (1) k  [–6, –2]

(2) k  [–6, 2]

(3) k  [–2, 6]

(4) k  [2, 6]

Sol. Answer (4) 2sin2x = 8 – 2k + k sinx  2sin2x – ksinx + 2k – 8 = 0

sin x 

k  k 2  4(2)(2k  8) k  k 2  6k  64 k  (k  8)   4 4 4

sin x 

2k  8 ,2 4


94


1


2k  8 1 4

 –4  2k – 8  4  4  2k  12 

2 k  6

136. The number of solutions of the x  2tan x  (1) 1

 in [0, 2] is 2

(2) 2

(3) 3

(4) 6

Sol. Answer (3)

2tan x 

 x 2

– 2

 x tan x   4 2

O

 2



3 2

2

3 solutions exist. 137. How many solutions does the equation sec x  1  ( 2  1)tan x have in the interval (0, 6]? (1) 6

(2) 5

(3) 10

(4) 9

Sol. Answer (1)

sec x  1 ( 2  1)tan x 1 cos x ( 2  1)sin x  cos x cos x  ( 2  1)sin x  cos x  1 Using auxiliary form, 6 solutions exist.

138. The solution of the equation

(1) x  3n 

 6

sin3 x 1  is 2cos2 x  1 2

(2) x  n 

 6

n (3) x  n  ( 1)

 3

n (4) x  n  ( 1)

 6

Sol. Answer (4)

sin3 x 1  2cos2 x  1 2  2(3sinx – 4sin3x) = 2(1 – 2sin2x) + 1 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



95

 6sinx – 8sin3x = 2 – 4sin2x + 1  8sin3x – 4sin2x – 6sinx + 3 = 0  4sin2x(2sinx – 1) – 3(2sinx – 1) = 0 sin x 

1 3 or sin2 x  2 4

x  n  ( 1)n

 6

139. The smallest positive values of x (in degrees) such that tan(x + 100°) = tan(x + 50°)tanx·tan(x – 50°) is equal to zero (1) 60°

(2) 30°

(3) 40°

(4) 50°

Sol. Answer (2)

tan( x  100)  tan( x  50).tan x tan( x  50) 

2sin( x  100)cos( x  50) 2sin( x  50)sin x  2sin( x  50)cos( x  100) 2cos( x  50)cos x



sin(2 x  50)  sin150 cos50  cos(2 x  50)  sin(2 x  50) – sin150 cos(2 x  50)  cos50



2sin(2 x  50) 2cos50  1 2cos(2 x  50)

 2sin(2x + 50°) cos(2x + 50°) = – cos50°  Sin(4x + 100°) = – sin40° = sin 220°,sin320°  4x + 100 = 220, 320  4x = 120, 220  x = 30, 55 Properties of Triangle 140. If the lengths of the sides of a triangle are 3, 5 and 7, then the largest angle of the triangle is (1)

 2

(2)

5 6

(3)

2 3

(4)

3 4

Sol. Answer (3)

141. In ABC,

bc = a

⎛ B C ⎞ cos ⎜ ⎟ ⎝ 2 ⎠ (1) A sin 2

⎛ B C ⎞ sin ⎜ ⎟ ⎝ 2 ⎠ (2) A cos 2

⎛B C ⎞ cos ⎜ ⎟ ⎝ 2 ⎠ (3) A sin 2

⎛B C ⎞ cos ⎜ ⎟ ⎝ 2 ⎠ (4) A cos 2

Sol. Answer (1) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

96



142. If a cosA = b cosB, then ABC is (1) Isosceles

(2) Right angled

(3) Equilateral

(4) Either Right angled or Isoceles

Sol. Answer (4)

⎛B⎞ 143. If a = 16, b = 24 and c = 20, then the value of cos ⎜ ⎟ is ⎝2⎠ (1)

3 4

(2)

1 4

(3)

1 2

(4)

1 3

Sol. Answer (1) Find cosB and then cos2 = 2cos2 –1 144. If a  1  3, b  2 and c  2, then the measure of B is (1) 45°

(2) 60°

(3) 90°

(4) 15°

(3) 2 3

(4)

Sol. Answer (1) 145. If A = 60°, b = 2 and c = 4, then the value of a is (1) 2 2

(2) 3 2

6

Sol. Answer (3) 146. If b  3, c = 1 and A = 30°, then the measure of B is (1) 30°

(2) 120°

(3) 45°

(4) 150°

Sol. Answer (2)

147. If in a triangle ABC,

sin A sin(A  B ) = , then sin C sin(B  C )

(1) a, b, c are in A.P.

(2) a2, b2, c2 are in A.P.

(3) a, b, c are in H.P.

(4) a2, b2, c2 are in H.P.

Sol. Answer (2) Given that

sin A sin( A  B )  sin C sin(B  C )  sinA.sin(B – C) = sinC. sin(A – B)  sin(B + C). sin(B – C) = sin(A + B). sin(A – B) [∵ A + B + C = 180°]  sin2B – sin2C = sin2A – sin2B  

b2 4R 2



c2 4R 2



a2 4R 2



b2 4R 2

2b2 = a2 + c2

 a2, b2, c2 are in A.P. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


148. In a ABC, if


97

cos A cos B cos C = = and the side a = 2, then area of the triangle is a b c

(1) 1

(2) 2

(3)

3 2

(4)

3

Sol. Answer (4) Given that ;

cos A cos B cos C   ⇒ A  B  C  60 a b c

So, area of the triangle

149. If cot

=

3 3 (side)2  4 4 4

=

3

A b+c = , then ABC is 2 a

(1) Isosceles

(2) Equilateral

(3) Right angled

(4) Scalene

Sol. Answer (3) Given that

cot

A bc  2 a

A 2  sin B  sin C  A sin A sin 2 cos

A B C B C 2 sin .cos 2  2 2  A A A sin 2 sin . cos 2 2 2 cos

A A B C 2cos .cos 2 2 2   A A A sin 2sin .cos 2 2 2 cos

A A B C 2cos .cos 2 2 2   A A A sin 2sin .cos 2 2 2 cos

 cos

(B  C ) A  cos 2 2

B – C = A A + C = B  But A + B + C = 180° 2B = 180°  B = 90° So, triangle is right angled triangle Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

98



⎛ AB ⎞ 3 a 7 = 150. In triangle ABC, if tan ⎜ and = then the value of angle C is ⎟ b 4 ⎝ 2 ⎠ 11

(1) 30°

(2) 60°

(3) 90°

(4) 45°

Sol. Answer (3) 3 a 7 ab 3 ⎛ A  B⎞  and Given that tan ⎜  ⇒  ⎟ ⎝ 2 ⎠ 11 b 4 a  b 11

We know that

AB ab C   cot 2 2 ab

C C 3 3  .cot ⇒ cot  1 ⇒ C  90 11 11 2 2

151. In a triangle, the length of two larger sides are 24 and 22 respectively. If the angles are in AP, then the third side is (1) 12 + 2 13

(2) 12 – 3 3

(3)

2 3 2

(4)

2 3 2

Sol. Answer (1) Given that, angles A, B and C are in A.P  B = 60° We know that cos B 



a2  c 2  b2 2ac

1 (24)2  c 2  (22)2  ⇒ c 2  92  24c 2 2  24  c

 c2 – 24c + 92 = 0  c  12  2 13 152. In a triangle ABC, a = 4, b = 3, A = 60º, then c is the root of the equation (1) c2 – 3c – 7 = 0

(2) c2 + 3c + 7 = 0

(3) c2 – 3c + 7 = 0

(4) c2 + 3c – 7 = 0

Sol. Answer (1) We know that

cos A 



b2  c 2  a2 2bc

1 9  c 2  16  ⇒ c 2  3c  7  0 2 23c

153. In triangle ABC, point D lies on BC such that BAD = 45°, DAC = 30°. If BD : DC = 2 : 1, then 3cotADC is equal to (1) 2 + 3

(2) 2 – 3

(3)

3

(4)

1 3




99

A

Sol. Answer (2) Given that BD : DC = 2 : 1

45° 30°

So from (m, n) theorem (2 + 1)cot = 2cot45 – 1.cot30



B

3 cot = 2  3

154. If in a ABC, 3a = b + c, then the value of cot (1) 1

(2)

C

D

B C .cot is 2 2

3

(3) 2

(4) 3

(3) b, a, c are in A.P.

(4) a, b, c are in G.P.

Sol. Answer (3) Given that 3a = b + c Now, cot

B C .cot  2 2

155. In ABC, tan

s (s  b ) s (s  c )  (s  a )(s  c ) (s  a)(s  b)

=

s s a

=

abc [∵ 2s = a + b + c] bc a

=

4a 2 2a

A 5 C 2 = , tan = , then 2 6 2 5

(1) a, b, c are in A.P.

(2) a, c, b are in A.P.

Sol. Answer (1) Given that tan

So, tan

A 5 c 2  , tan  2 6 2 5

A C 5 2 ,tan   2 2 6 5

(s  b )(s  c ) (s  a)(s  b) 1   s (s  a ) s (s  c ) 3



sb 1 ac b 1  ⇒  s 3 abc 3

2b = a + c Hence a, b, c are in A.P. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

100



156. If in a triangle ABC,  = a2 – (b – c)2, then tanA is equal to (1)

15 16

(2)

8 15

(3)

8 17

(4)

1 2

Sol. Answer (2) Given that  = a2 – (b – c)2 s(s  a )(s  b )(s  c )  4(s  b )(s  c )

⎡ A ⎢∵ tan  2 ⎣⎢

A (s  b )(s  c ) 1   tan s (s  a ) 4 2

(s  b )(s  c ) ⎤ ⎥ s(s  a ) ⎦⎥

1 A 2 4  8 2 = Now, tan A  1 15 A 1 1  tan2 16 2 2 tan

 tan A 

8 15





157. If the angles of a triangle are 30° and 45° and the included side is 1+ 3 cm , then the area of the triangle is (in square cm)



(1) 2 1 + 3



(2)



1 1+ 3 2



(3) 2





3 1

(4)

1 2





3 1

Sol. Answer (2)

A c

B

105°

b

30°

45° a=1+ 3

C

From Sine Rule

1 3 c  ⇒c 2 sin105 sin 45

Hence Area =

1 1 ac sin B = (1  3 ) 2 2

158. If c2 = a2 + b2, 2s = a + b + c, then 4s(s – a)(s – b)(s – c) (1) s4

(2) b2c2

(3) c2a2

(4) a2b2




101

Sol. Answer (4)

B

Given that c2 = a2 + b2  c = 90°

c

Now,

a

4s(s – a) (s – b) (s – c) = 4(Area)2 = 4

A

1 2 2 a b  a2 b2 4

C

b

159. A triangle has its sides in the ratio 4 : 5 : 6, then the ratio of circumradius to the inradius of the triangle is (1)

15 7

(2)

13 6

(3)

16 7

(4)

17 6

Sol. Answer (3) Let a = 4, b = 5, c = 6 We know that

r R

cosA + cosB + cosC = 1 

b2  c 2  a2 a2  c 2  b2 a2  b2  c 2 r    1 2bc 2ac 2ab R 1

r 23 r 7 R 16  ⇒  ⇒  R 16 R 16 r 7

160. In ABC, if ar(ABC) = 8. Then, a2sin(2B) + b2sin(2A) is equal to (1) 2

(2) 16

(3) 32

(4) 128

Sol. Answer (3) Given that  = 8 Now, a2sin2B + b2sin2A = 2a2sinB cosB + 2b2sinAcosA 2 = 2a 

=

b a .cos B  2b 2   cos A 2R 2R

ab (a cos B  b cos A) R

abc ⎛ abc ⎞ = R  4 ⎜⎝ 4R ⎟⎠  4   32

 a2 sin2B  b2 sin2 A  32 161. In ABC, if a2 + b2 + c2 = ac + ab 3 , then

r r3 (1) r = r 2 1

r r2 (2) r = r 1 3

(3) r1 + r2 + r3 = 2R + r

(4) rr1r2r3 = 43


102



Sol. Answer (1) Given that, a2 + b2 + c2 = ac + ab 3  a2 + b2 + c2 – ac – ab 3 = 0 2

2 ⎛a 3 ⎞ ⎛a ⎞  b⎟  ⎜  c ⎟  0  ⎜ ⎝2 ⎠ ⎝ 2 ⎠

Above equation is possible when

a 3 a b  0& c  0 2 2

3a  2b  2 3c  k (let) a 

k 3

,b 

k k ,c  2 2 3

 b2 + c2 = a2 triangle is right angled triangle now taking (1) option r3 r  r  r ⇒ rr1  r2 r3 2 1



2 2  s(s  a ) (s  b )(s  c )

 s2 – sa = s2 – s(b + c) + bc  (b + c – a) (b + c + a) = 2bc  (b + c – a)s – bc  (b + c)2 – a2 = 2bc  b2 + c2 – a2 = 0 

b2  c 2  a2 proved

So option (1) is correct 162. In a triangle ABC, ab and c are the sides of the triangle satisfying the relation r1 + r2 = r3 – r then the perimeter of the triangle (1)

2ab abc

(2)

ab ac b

(3)

ab abc

(4)

bc bc a

Sol. Answer (1) Given that r1 + r2 = r3 – r 

       s a s b s c s

(s – a) (s – b) = s(s – c)



2s  a  b s sc  (s  a )(s  b ) s(s  c )

 2s 

2ab abc

Hence, option (1) is correct Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



103

163. In ABC, which is not right angled, if p = sinA sinB sinC and q = cosA cosB cosC. Then the equation having roots tanA, tanB and tanC is (1) qx3 – px2 + (1 + q)x – p = 0

(2) qx3 + 2px2 + qx – p = 0

(3) qx3 – px2 + (2 + q)x + pq = 0

(4) qx3 – px2 + qx + p = 0

Sol. Answer (1) Given that p = sinA.sinB.sinC q = cosA. cosB. cosC Now, tanA. tanB. tanC =

p q

And also tanA + tanB + tanC = tanA.tanB.tanC =

p q

Now, tanA.tanB + tanB.tanC. + tanC.tanA =

sin A.sin B sin B.sin C sin C.sin A   cos A.cos B cos B.cos C cos C.cos A

= sinA.sinB.cosC + sinBsinC.cosA + sinAsinC.cosB/cosAcosBcosC = sinB(sinAcosC + cosAsinC) + sinA.sinC.cosB/cosAcosBcosC = sinB sin(A + C) + sinA. sinC . cosB /cosAcosBcosC =

sin2 B  sin A sinC.cos B cos A.cos B.cos C

=

1  cos2 B  sin A.sin C.cos B cos A.cos B.cos C

=

1  cos B(cos B  sin A sin C ) cos A.cos B.cos C

=

1  cos B {  cos( A  C )  sin A sin C } cos A.cos B.cos C

=

1  cos B.cos A cos C 1  q  q cos A.cos B.cos C

So required equation is

x3 

p 2 ⎛ 1 q ⎞ p x ⎜ x 0 ⎟ q q ⎝ q ⎠

 qx 3  px 2  (1  q )x  p  0 Hence option (1) is correct Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

104


164. In ABC, if


R  2 , and a = 2, then  is equal to ( = ar(ABC)) r

(1) 3

(2) 9

(3)

3

(4)

3 4

Sol. Answer (3) Given that R r 1 2⇒  r R 2

 1

r 3 3  ⇒ cos A  cos B  cos C  2 R 2

…(i)

But we know that cosA + cosB + cosC

3 2

…(ii)

So from (i) & (ii) we get

3 2

 cosA + cosB + cosC =  A=B=C Hence triangle is equilateral

3 (sinde)2 4

So area of the triangle is =

=

3 4 3 4

 Option (3) is correct ⎛ 1 1 ⎞ ⎛ 1 1 ⎞⎛ 1 1 ⎞ 165. In ABC, if r1, r2, r3 are exradii opposites to angles A, B and C respectively. Then ⎜ + ⎟ ⎜ + ⎟⎜ + ⎟ ⎝ r1 r2 ⎠⎝ r2 r3 ⎠⎝ r3 r1 ⎠

is equal to

(1)

64R 3 abc

(2)

16R 3 2 2 2

a b c

(3)

64R 3 2 2 2

a b c

(4)

R3 abc

Sol. Answer (3) ⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1⎞ ⎜⎝ r  r ⎟⎠ ⎜⎝ r  r ⎟⎠ ⎜⎝ r  r ⎟⎠ 1 2 2 3 3 1

=

=

abc 3

64R 3 ⎛ using abc  R ⎞ ⎜ ⎟⎠ 4 a2 b2c 2 ⎝

So option (3) is correct. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


166.


105

2cos A cos B 2cos C a b + + = + , then the value of the angle A is a b c bc ca  3

(1)

(2)

 4

(3)

 2

(4)

 6

Sol. Answer (3)

2cos A cos B 2cos C a b     a b c bc ca

2(b2  c 2  a2 ) a2  c 2  b2 2(a2  b2  c 2 ) a2  b2    2abc 2abc 2abc abc



 a2 + 3b2 + c2 = 2a2 + 2b2  c2 + b2 = a2  ABC is right angled  and A is a right angle. So option (3) is correct 167. In triangle ABC if the line joining incentre to the circumcentre is parallel to the base BC, then the value of (cosB + cosC – 1) is equal to (1) 1

(2) 2

(3) 3

(4) 0

Sol. Answer (4)

A

In ABC cosA + cosB + cosC = 1 

r R

I

r r  cos B  cos C  1   R R

R B

 cosB + cosC – 1 = 0

r

O A r

cosA = r

C

R

So, option (4) is correct 168. With usual notations, if in a triangle ABC, b+c c+a a+b = = , then cosA : cosB : cosC is equal to 11 12 13

(1) 7 : 19 : 25

(2) 19 : 7 : 25

(3) 12 : 14 : 20

(4) 19 : 25 : 20

Sol. Answer (1) Let

bc ca ab   k 11 12 13

 b + c = 11 k c + a = 12 k a + b = 13 k 2(a + b + c) = 36 k a + b + c = 18 k  a = 7 k, b = 6 k, c = 5 k Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

106



cosA : cosB : cosC

b2  c 2  a2 a2  c 2  b2 a2  b2  c 2 : : 2bc 2ac 2ab Put the values of a, b & c 7 : 19 : 25 So option (1) is correct 169. Consider a regular polygon of 12 sides each of length one unit, then which of the following is not true?



(1) The area of polygon is 6 + 3 3

(3) The radius of incircle is



(2) The circum-radius of polygon is

2+ 3 2

1 2



6+ 2



(4) Each internal angle is of 135°

Sol. Answer (4) A

 AOP  12

=

=

=

O

1  AB  OP 2

= 6  1 OP

B

11 11 12 12

Area of polygon = 12  ArAOB = 12 

P

tan  / 12 

1 2OP

6 2 tan  / 12 3 tan  / 12

3 3 1

( 3  1)

= 63 3 So option (1) is true (ii) sin

 1  12 2R

 R 

 R 

A

P 1/2 /12

1

 2 sin 12

B

R O

6 2 2




 1  12 2r

r

/12

1/2

So option (2) is true

tan

107

2 3 2

 r 

So option (3) is true (iv) Each internal angle = =

(2n  4)90 n 20  90 = 150° 12

Hence option (4) is answer not true 170. If the perimeter of a triangle is 6, then its maximum area is (1)

3

(2) 3

(3) 4

(4) 2

Sol. Answer (1)

a

b

a+b+c=6 c

(s  a )  (s  b )  ( b  c )  {(s  a )(s  b )(s  c )}1/3 3 s ⎛ s(s  a )(s  b )(s  c ) ⎞  ⎟⎠ 3 ⎜⎝ s

s ⎛ A2 ⎞  3 ⎜⎝ s ⎟⎠

1/3

1/3

s A2/3  3 s1/3

s 4/3  A2/3 3 ⎛ s 4/3 ⎞ ⎜ 3 ⎟ ⎝ ⎠ s2 3 3

3/2

A

A

 max A  3 Hence option(1) is correct Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

108



171. In a triangle ABC, if A – B = 120° and R = 8r, then the value of cosC is (1)

7 8

(2)

3 4

(3)

4 5

(4)

3 5

Sol. Answer (1) A – B = 120° R = 8r r  4R sin

A B C sin sin 2 2 2

R A B C  4R sin sin sin 8 2 2 2 AB A  B⎞ C 1 ⎛  ⎜ cos  cos sin ⎟ 16 ⎝ 2 2 ⎠ 2 A  B⎞ C 1 ⎛1   cos sin 16 ⎜⎝ 2 2 ⎟⎠ 2 C⎞ C 1 ⎛1   sin ⎟ sin 16 ⎜⎝ 2 2⎠ 2 1 ⎛1 ⎞  ⎜  x⎟ x ⎝ ⎠ 16 2

Let sin

C x 2

 16x2 – 8x + 1 = 0

 x 

 sin

1 4

C 1  2 4

2  cos C  1  2 sin C / 2 

7 8

Hence option (1) is correct 172. If in a triangle 2(acosB + b cosC + c cosA) = a + b + c, then (1) The triangle is isosceles

(2) The triangle is equilateral

(3) The triangle is isosceles right angled

(4) The triangle is right angled

Sol. Answer (1) 2[acosB + bcosC + C cosA] = a + b + c  2[2RsinA cosB + 2RsinBcosC + 2RsinCcosA] = a + b + c  2R[2sinA cosB + 2sinB cosC + 2sinC cosA] = 2R(sinA + sinB + sinC)  sin[A + B) + sin(A – B) + sin(B + C) + sin(B – C) + sin(C – A) + sin(C + A) = sinA + sinB + sinC Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



109

sin(A – B) + sin(B – C) + sin(S – A) = 0  sin

A C C B BA sin sin 0 2 2 2

Which is only possible when A = B or B = C or A = C  Triangle is isosceles Hence option (1) is correct 173. A regular pentagon is inscribed in a circle. If A1 and A2 represents the area of circle and that of regular pentagon respectively, then A1 : A2 is (1)

 2 cos 5 5

(2)

2 2 sin 5 5

(3)

2 2 cosec 5 5

(4)

2 2 cos 5 5

Sol. Answer (3) A1  A2

A

a

r 2 1 5 ah 2

E

/5 r

r r = 5 ah 2

=

2  5

1

 2 sin 5

B O 

1 cos

 5

D C

2 1   5 = sin 2 5

 Option (3) is true 174. In ABC, if 8R2 – a2– b2 = c2 then the triangle must be (1) Equilateral

(2) Right angle

(3) Scalene

(4) Isoceles angle

Sol. Answer (2) 175. In ABC, if O is the circumcentre of ABC and R1, R2, R3 are the radii of circumcircles of triangles OBC, OCA and OAB respectively. Then the value of

(1)

abc R

(2)

abc 2R

(3)

A

Sol. Answer (4) Let AOB = 2

c

2

2

abc 4R

3

(4)

abc R3

b

2

BOC = 2

B AOC = 2

a b c + + is equal to R1 R2 R3

a

C


110



In AOB

c  2R3 sin2 C  2sin2 R3

a Similarly R  2sin2 1 b  2sin2 R2

a b c  R  R  R 1 2 3 = 2 sin2 + 2sin2 + 2sin2 = 2[sin2 + sin2 + sin2] = 2[4sin sin sin] = 8sin sin sin = 8

=

a b c   2R 2R 2R

abc R3

Hence option (4) is correct 176. In ABC, if r1 = 2r2 = 3r3, then (1)

a 5 = b 4

(2)

a 3 = b 5

(3)

a 7 = c 4

(4)

c 4 = a 9

Sol. Answer (1) r1 = 2r2 = 3r3

 2 3   s a s b s c 1 2 3   s a s b s c Taking

1 2  s a s b

 3a = 3b + c Taking

…(i)

1 3  s a s c




 2a = b + 2c

111

…(ii)

Solving (i) & (ii) a 5  b 4

Hence option (1) is correct 177. In ABC, the expression b2  c 2 c 2  a2 a2  b2 + + is equal to a sin(B  C ) b sin(C  A) c sin(A  B )

(1)

3R 2

(2) 3R

(3) R

(4) 6R

Sol. Answer (4) b2  c 2 (2R sin B )2  (2R sin C )2  a sin(B  C ) 2R sin A sin(B  C )

=

4R 2 (sin2 B  sin2 C ) 2R sin A sin(B  C )

=

4R 2 sin(B  C )sin(B  C ) 2R sin A sin(B  C )

= 2R other terms are also 2R  2R + 2R + 2R = 6R

178. In a triangle ABC if  2 =

Hence option (4) a2 b2c 2 2(a 2 + b 2 + c 2 )

, where  is the area of the triangle, then the triangle is

(1) Isosceles but not right angled

(2) Right angled

(3) Isosceles right angled

(4) Equilateral

Sol. Answer (2) 2 Given that  

a2 b2c 2 2(a2  b 2  c 2 )

But we know that

2 

abc  R ⇒ abc  4R 4

16 2R 2 2

2

2

2(a  b  c )

⇒ a2  b2  c 2  8R 2

 sin2A + sin2B + sin2C = 2

…(i)

From (i) it is clear that the triangle will be right-angled triangle. So, option (2) is correct Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

112



179. ABCD is a trapezium such that AB and DC are parallel and BC is perpendicular to them. If BC = 1 cm, CD = 2 cm and ADB = 45°, then the length of AB is (in cm) (1)

12 5

(2)

8 5

(3)

Sol. Answer (3)

D

Using cosine formula in DCB cos  

5  4 1 4 5



2 5

A



2

5 3

(4)

16 3

C

5

1

135 –

B

Apply law of sin in ADB sin 45 sin(135 –  )  AB 5





1 2 AB



sin135 cos   cos135 sin  5

⎛ 2 1 ⎞ 1 ⎜  ⎟ 2 AB ⎝ 10 10 ⎠ 5 1

 AB 

5 3

Here option (3) is correct 180. If the median of triangle ABC through A is perpendicular AB, then the value of sinA cosB + 2sinB cosA is equal to (1) 0

(2) 1

(3)

1 2

(4)

1 3

A

Sol. Answer (1) From (m1 n) theorem 2x cot(90 + B) = xcot0 – xcot(A – 90) 

A–90°

– 2tanB = tanA

 – 2sinB cosA = sinAcosB  sinAcosB + 2sinBcosA = 0  Hence option (1) is correct

90° + B B

x

D

x

C

181. The number of solutions of the pair of equations 2 sin2 – cos2 = 0 2 cos2 – 3 sin = 0 in the interval [0, 2] is (1) Zero

(2) One

[IIT-JEE 2007] (3) Two

(4) Four




113

Sol. Answer (3) 1 4

2sin2 – cos2 = 0  sin2 = 2cos2 – 3sin = 0  sin =

1 –2 2

common solutions are given by sin =

1 2

182. If the angles A, B and C of a triangle are in an arithmetic progression and if a, b and c denote the lengths of the sides opposite to A, B and C respectively, then the value of the expression

(1)

1 2

(2)

3 2

(3) 1

a c sin 2C  sin 2 A is [IIT-JEE 2010] c a

(4)

3

Sol. Answer (4) Angles A, B, C are in A.P. hence B = 60° =

 3

⎛ sin C ⎞ ⎛ sin A ⎞ a c a cos C  2 ⎜ sin 2C  sin 2 A = 2 ⎜ ⎟ ⎟ c cos A c a ⎝ c ⎠ ⎝ a ⎠ = 2(sinA cos C + cosA sinC) = 2 sin(A + C) = 2 sinB = 2 

3  3 2

183. Let P  { : sin   cos   2 cos } and Q  { : sin   cos   2 sin } be two sets. Then (1) P Q and Q – P 

(2) Q P

(3) P Q

[IIT-JEE 2011]

(4) P = Q

Sol. Answer (4) We have sin   cos   2 cos 

 sin   ( 2  1)cos  

1 2 1

sin   cos 

 ( 2  1)sin   cos   2 sin   sin   cos  and it shows that the given equations  sin   cos   2 cos  and sin   cos   2 sin  are identical. Hence their solution sets are equal  P=Q Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

114



184. Let PQR be a triangle of area  with a = 2, b =

7 5 and c = , where a, b and c are the lengths of the sides of 2 2

the triangle opposite to the angles at P, Q and R respectively. Then

(1)

3 4

(2)

⎛ 3 ⎞ (3) ⎜ ⎟ ⎝ 4 ⎠

45 4

2 sin P  sin 2P equals 2 sin P  sin 2P 2

⎛ 45 ⎞ (4) ⎜ ⎟ ⎝ 4 ⎠

[IIT-JEE 2012] 2

Sol. Answer (3) As,

2sin P – sin2P 1 – cos P ⎛P ⎞   tan2 ⎜ ⎟ 2sin P  sin2P 1  cos P ⎝2⎠

Now, a = 2, b 

7 5 , c 2 2

so, semi perimeter = s 

Now, tan

abc 4 2

P (s – b )(s – c ) 3   2  4

⎛ 3 ⎞ 2⎛P ⎞ So, tan ⎜ ⎟  ⎜ ⎟ ⎝ 2 ⎠ ⎝ 4 ⎠

2

185. In a triangle the sum of two sides is x and the product of the same two sides is y. If x2 – c2 = y, where c is the third side of the triangle, then the ratio of the in radius to the circum-radius of the triangle is [JEE(Advanced)-2014 ] (1)

3y 2x( x  c )

(2)

3y 2c ( x  c )

(3)

3y 4 x( x  c )

(4)

3y 4c ( x  c )

Sol. Answer (2) a + b = x, ab = y, x2 – c2 = y Now, (a + b)2 – c2 = ab a2 + b2 – c2 = –ab a2  b2  c 2 1  2 ab 2

 cos c  

2R 

1 2

c = 120°

c 2c  sin c 3

r  (s  c )  tan

 R 

c 3

c 2




115

⎛x c ⎞ r ⎜ c⎟ 3 2 ⎝ ⎠ r 

(x  c) 3 2

r (x  c) 3 3( x  c )   3 R 2c 2c 

3( x  c )( x  c ) 2 c( x  c )



3( x 2  c 2 ) 2 c( x  c )



3y 2 c( x  c )

186. For x  (0, ), the equation sinx + 2sin2x – sin3x = 3 has (1) Infinitely many solutions

(2) Three solutions

(3) One solution

(4) No solution

[JEE(Advanced)2014]

Sol. Answer (4) sinx + 2sin2x – sin3x = 3 2cos2x(–sinx) + 4sinxcosx = 3 2sinx[2cosx – cos2x] = 3 2sinx(2cosx – (2cos2x – 1)) = 3 2sinx(1 + 2cosx – 2cos2x) = 3 2 ⎡3 1⎞ ⎤ ⎛ 2 sin x ⎢  2 ⎜ cos x  ⎟ ⎥  3 2⎠ ⎥ ⎝ ⎢⎣ 2 ⎦

Possible only when sinx = 1

…(i) 2

1⎞ ⎛ and ⎜ cos x  ⎟  0 2⎠ ⎝  cos x 

1 2

…(ii)

From (i) and (ii) No solution. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

116



SECTION - B Objective Type Questions (More than one options are correct) Elementary Trigonometric Functions 1.

If x  sec   tan  and y  cosec  cot  , then

(1) x 

y 1 y 1

(2) x 

y 1 y 1

(3) y 

1 x 1 x

(4) xy  x  y  1 0

Sol. Answer (2, 3, 4) x = sec – tan, y = cosec + cot y  cot

 2 2

 ⎞ ⎛ cos  sin ⎟ 1 sin  ⎜⎝ 2 2⎠  x   cos  cos2  sin2 2 2 2

 ⎞ ⎛ ⎜ cot 2  1⎟ 2 ⎠  ( y  1)  y  1  x  ⎝ 2  y 1 cot 2  1 ( y  1) 2

2.

⎧  3 ⎫ The value of F ( x )  6cos x 1 tan2 x  2sin x 1 cot  x where x (0, 2)  ⎨, , ⎬ may be ⎩ 2 2⎭ (1) 4

(2) – 4

(3) 8

(4) – 8

Sol. Answer (1, 2, 3, 4) 3 ⎫ ⎧ f ( x )  6 cos x 1  tan2 x  2 sin x 1  cot 2 x , x  (0, 2)  ⎨ , , ⎬ 2 2 ⎭ ⎩

= 6cosx. |secx| + 2sinx. |cosecx| ⎧ ⎛ ⎞ x  ⎜ 0, ⎟ ⎪ 628 ⎝ 2⎠ ⎪ ⎪ ⎛ ⎞ x  ⎜ , ⎟ ⎪6  2  4 ⎪ ⎝2 ⎠ ⎨ ⎪ 6  2  8 x  ⎛ , 3 ⎞ ⎜ 2 ⎟ ⎪ ⎝ ⎠ ⎪ ⎛ 3 ⎞ ⎪ x  ⎜ , 2 ⎟ ⎪ 62  4 2 ⎝ ⎠ ⎩ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


3.

If


117

(a  b )2  sin2  , then 4ab (2) a  0, b  0

(1) a = b

(3) a > b

(4) a < b

Sol. Answer (1, 2) (a  b )2  sin2  4ab

∵ 0 sin2 1 

(a  b )2 1 4ab

 (a – b)2  0  a = b, but not equal to zero

4.

If sec   x 

1 , then tan   sec  is equal to 4x

(1) 2x

(2)

1 2x

(3)

1 x

(4) 4x

Sol. Answer (1, 2) sec = x 

1 4x

tan   sec 2   1 2

=

1 ⎞ ⎛ ⎜ x  4x ⎟  1 ⎝ ⎠

= x

1 4x

1 ⎞ 1 ⎛  x  sec + tan = ⎜ x  ⎟ 4x ⎠ 4x ⎝

⎧ ⎪⎪ 2 x = ⎨ ⎪ 1 ⎪⎩ 2 x

5.

1 4x 1 if x  4x

if x 

If tan   cot   2, then (1) tann   cot n   2 (3) sin2n   cos2n  

(2) 1 2n 1

tan   cot   2

(4) tan2   sec 2   3


118



Sol. Answer (1, 2, 3, 4) tan + cot = 2  tan2 + 1 = 2tan  tan = 1 (1) tan4 + cot4 = 2 tan   cot   2

(2)

⎛ 1 ⎞ 2n 2n (3) sin   cos   ⎜  ⎟ 2⎠ ⎝

=

2n

⎛ 1 ⎞  ⎜ ⎟ 2⎠ ⎝

2n

1 2

n1

(4) tan2 + sec2 = 1 + 2 = 3 6.

Which of the following identities are correct? (1)

1 sin  cos    2 sec  cos  1 sin 

(3)

1 sin  | sec   tan  | 1 sin 

(2) (1  cot A  cos ecA)(1  tan A  sec A)  2

(4)

cos  cos    2 sec  1 sin  1 sin 

Sol. Answer (1, 2, 3, 4) (1)

1  sin  cos    2 sin  cos  1  sin 

L.H.S 

1  sin  cos (1 – sin )  cos  cos2  2  RHS cos 

=

(2) (1 + cotA – cosecA) (1 + tanA + secA) = 2 ⎛ sin A  cos A  1 ⎞ ⎛ cos A  sin A  1 ⎞ L.H.S. = ⎜ ⎟⎜ ⎟ sin A cos A ⎝ ⎠⎝ ⎠

=

(3) =

1  2 sin A cos A  1  2  R.H.S sin A  cos A

1  sin   | sec   tan  | 1  sin 

L.H.S. =

(1  sin )2 cos2 

= |sec + tan| = R.H.S. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


(4)


119

cos  cos    2 sec  1  sin  1  sin 

L.H.S. 

=

cos(1  sin ) cos2 

cos   (1  sin )



cos2 

1  sin  1  sin   cos  cos 

= 2sec = R.H.S 7.

The value of the expression sin x  sin y  sin z where x, y, z are real numbers satisfying x + y + z = 180° is (1) Positive

(2) Zero

(3) –3

(4) Negative

Sol. Answer (1, 2, 4) x + y + z = 180° x = –90°, y = 90°, z = 360°

8.

⎛ ⎞ Let  ⎜ 0, ⎟ and a  (tan )tan , b  (tan )co t  , c  (cot )tan , d  (cot )cot  , then ⎝ 4⎠ (1) d > c

(2) b < a

(3) c > a

(4) b < a < c < d

Sol. Answer (1, 2, 3, 4) ⎛ ⎞   ⎜ 0, ⎟ ⎝ 4⎠

 0 < tan < 1, 1 < cot <  a = (tan)tan, b = (tan)cot c = (cot)tan, d = (cot)cot Clearly b is the smallest and d is the greatest of all Also, Fractional value (number greater than 1) > 1  b
⎛ ⎞ ⎛ ⎞ If tan ⎜ + x ⎟ ·tan ⎜  x ⎟ = 1 , then x equals ⎝4 ⎠ ⎝4 ⎠ (1)

 6

(2)

 3

(3)

 4

(4) Zero

Sol. Answer (1, 2, 4)

⎛ ⎞ ⎛ ⎞ tan ⎜  x ⎟ .tan ⎜  x ⎟  1 ⎝4 ⎠ ⎝4 ⎠ 

1  tan x 1  tan x  1 1  tan x 1  tan x


120



1 10. The value of cos 22  is 2

(1)

1 2 2 2

(3)

1 4

 4+2 2 +

42 2



 4+2 2 

(2)

1 4

(4)

1 2 2 2

42 2



Sol. Answer (1, 3)

1 cos 22   ? 2 2cos2 = 1 + cos2 1 1  cos 45 ⇒ cos 22    2 2

2 2  4





cos

 1  2 2



1 1 ⇒ cos 22   2 2

2 2 2

1  sin   1  sin 



1⎛ 1 1 ⎞  1 ⎜⎜ 1  ⎟ 2⎝ 2 2 ⎟⎠



1⎛ ⎜ 2⎜ ⎝



1⎛ 4  2 2 42 2 ⎞ ⎜ ⎟  ⎟ 2⎜ 4 4 ⎝ ⎠



1 4



2

2 tan



2 1⎞ ⎟ 2 ⎟⎠

42 2  42 2

11. Let cos 2 =

(1)





1  sin 45  1  sin 45



2 1

21 2 2



3cos 2β  1 , then tan is equal to 3  cos 2β (2)  2 tan 

(3)

1 2

tan 

(4) 

1 2

tan 




121

Sol. Answer (1, 2)

3cos 2  1 3  cos 2

cos 2 

1  tan2 



2

1  tan 

3 cos 2  1 3  cos 2

1  tan2   1  tan2  2

2

1  tan   1  tan 

2 2

2 tan  

1 tan2 





3 cos 2  1  3  cos 2 3 cos 2  1  3  cos 2

2cos 2  2 4cos 2  4

2(1  cos 2) 2cos2   4(1  cos 2) 2  2 tan2 



tan2 = 2tan2 tan =  2 tan 

12. If sin   

(1)

 3 ,   [0, 2] then the possible values of cos is/are 2 5

1

(2) 

10

1 10

(3)

3 10

(4) 

3 10

Sol. Answer (2, 4) sin   

cos

3 5

 1  2 2



1  sin   1  sin 



1⎛ 3 3⎞ ⎜⎜ 1   1  ⎟⎟ 2⎝ 5 5⎠



1⎛ 2 2 2⎞  ⎜⎜ ⎟ 2⎝ 5 5 ⎟⎠



1⎛ 4 2 ⎞ 3  ⎜ ⎟ 2 ⎝ 10 10 ⎠ 10



13. The expression f(, ) = cos2 + cos2( + ) – 2coscoscos( + ) (1) Is independent of 

(2) Is independent of 

(3) Is independent of both  and 

(4) Is dependent of 


122



Sol. Answer (2, 4) f(, ) = cos2 + cos2( + ) – 2cos cos cos( + ) = cos2 + cos( + ) (cos cos – sin sin – 2cos cos) = cos2 – cos( + ) cos( – ) = cos2 – (cos2 – sin2) = cos2 – cos2 + sin2 = sin2 14. The value of cosA·cos2A·cos22A ........ cos(2n – 1A), where A R may be (1) 1

(2) 2

(3) –1

(4)

sin 2n A 2n sin A

Sol. Answer (4) cos A · cos2 A........cos(2n 1 A) 

sin2n A 2n sin A

 and cosA + cosB = 1, then 3

15. If A  B 

(1) cos( A  B ) 

1 3

(3) cos( A  B )  

(2) | cos A  cos B | 1 3

(4) | cos A  cos B |

2 3

1 2 3

Sol. Answer (2, 3) AB 

 , cos A  cos B  1 3

AB AB cos 1 2 2

⇒

2cos

⇒

AB ⎛⎞ 2cos ⎜ ⎟ cos 1 2 ⎝6⎠

⇒

2

⇒

1 ⎛ AB⎞ cos ⎜ ⎟ 3 ⎝ 2 ⎠



3 ⎛ AB⎞  cos ⎜ ⎟ 1 2 ⎝ 2 ⎠

⎛ AB⎞ ⎛ AB⎞ cos ⎜ ⎟  cos 2 ⎜ 2 ⎟ ⎝ 2 ⎠ ⎝ ⎠

⎛ AB⎞  2cos2 ⎜ ⎟ 1 ⎝ 2 ⎠  2 

1 1 3

2 1 1  3 3






1 ⎛ AB⎞ sin ⎜  1  ⎟ 3 ⎝ 2 ⎠

cos A  cos B  2 sin

123

2 3

AB ⎛ AB⎞ sin ⎜ ⎟ 2 ⎝ 2 ⎠

1 ⎛ AB⎞  2  sin ⎜ ⎟ 2 ⎝ 2 ⎠

⎛ AB⎞  sin ⎜ ⎟ ⎝ 2 ⎠ 

16.

2 3

cos3 x·sin2x =

n

∑am sin mx

is an identity in x, then

m =1

(1) a3 

3 , a2 = 0 8

(2) n = 6, a1 

1 2

(3) n = 5, a1 

1 4

(4)

3

∑ am  4

Sol. Answer (1, 3) cos3 x.sin 2 x 

n

∑ am sin mx

m 1

= a1sinx + a2sin2x + a3sin3x + ……. + an sinnx  2cos3x.cos2x – 3cos2xsin2x.sinx = a1cosx + 2a2cos2x + 3a3cos3x + ….. Put x = 0 2 = a1 + 2a2 + 3a3 + …… + nan Sn + 1 – Sn = a1 + 2a2 + …….. + nan + (n + 1)an + 1 – a1 – 2a2 ……. nan  En + 1 = (n + 1)an + 1  1 (cos3n  3cos n )sin2n 4



a1 



1 [2 sin 2 x cos3 x  3.sin 2 x sin x ] 8



1 [sin5 x  sin x  3 sin3 x  3 sin x ] 8



1 [sin5 x  3 sin3 x  2 sin x ] 8

1 3 1 , a2  0, a3  , a5  4 8 8


124



17. If 0 < ,  <  and they satisfy cos + cos – cos( + ) = (1)  = 

(2)  + =

2 3

3 2

(3)  = 2

(4)  = 2

Sol. Answer (1, 2) cos + cos – cos( + ) =  cos    

3 2

1 1 1 , cos   , cos(   )   2 2 2

  2 ,   ,   3 3 3

18. The angles A, B, C of a triangle ABC satisfy 4cosAcosB + sin2A + sin2B + sin2C = 4. Then which of the following statements is/are correct? (1) The triangle ABC is right angled (2) The triangle ABC is isosceles (3) The triangle ABC is neither isosceles nor right angled (4) The triangle ABC is equilateral Sol. Answer (1, 2) 4cosA.cosB + sin2A + sin2B + sin2C = 4

1 1 3 ⇒ 4. .  3 2 2 2 2[cos(A + B) + cos(A – B)] + 2cos(A + B) cos(A – B) + 2sinC cosC = 2  – cosC + cos(A – B) – cosC cos(A – B) + sinC cosC = 2  – cocC + cos(A – B) + cosC 19. Which of the following statement(s) is/are correct? (1) cos(sin1) > sin(cos1)

(2) cos(sin1.5) > sin(cos1.5)

7 ⎞ 7 ⎞ ⎛ ⎛ (3) cos ⎜ sin ⎟ > sin ⎜ cos ⎟ 18 ⎠ 18 ⎠ ⎝ ⎝

5 ⎞ 5 ⎞ ⎛ ⎛ (4) cos ⎜ sin ⎟ > sin ⎜ cos ⎟ 18 ⎠ 18 ⎠ ⎝ ⎝

Sol. Answer (1, 2, 3, 4)




125

20. If sin = K, –1 K 1, then number of values of , for same value of K in [0, 2] may be (1) 1

(2) 2

(3) 3

(4) 4


21. If the sides of a right angled triangle are {cos2 + cos2 + 2cos( + )} and {sin2 + sin2 + 2sin( + )}, then the length of hypoteneous is (1) 2[1 + cos( – )]

(2) 2[1 – cos(+ )]

⎛ ⎞ (3) 4cos2 ⎜ ⎟ ⎝ 2 ⎠

⎛⎞ (4) 4 sin2 ⎜ ⎟ ⎝ 2 ⎠

Sol. Answer (1, 3) a = cos2 + cos2 + 2cos( + ) = 2cos( + )cos( + ) + 2cos( + )

⎛ ⎞ = 2cos( + ) × 2cos2 ⎜⎝ 2 ⎟⎠ ⎛ ⎞ = 4cos( + )cos2 ⎜⎝ 2 ⎟⎠ b = sin2 + sin2 + 2sin( + ) = 2sin( + )cos( – ) + 2sin( + )

⎛ ⎞ = 2sin( + ).2cos2 ⎜⎝ 2 ⎟⎠

⎛ ⎞ = 4sin( + )cos2 ⎜⎝ 2 ⎟⎠ ⎛ ⎞ ⎛ ⎞ 2( + ) cos4 + 16sin2( + ).cos4 ⎜ ⎟  a2 + b2 ⎜ = 16cos ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎛ ⎞ = 16cos4 ⎜⎝ 2 ⎟⎠ ⎛ ⎞ A = 4cos2 ⎜⎝ 2 ⎟⎠ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

126



22. Let a and b be real numbers such that

sina + sinb =

1 2

, cosa + cosb =

(1) cos(a – b) = 1

3 , then 2

(2) cos(a – b) = 0

(3) sin(a + b) =

3 2

(4) sin(a – b) = 1

Sol. Answer (2, 4) 1 + 1 + 2cos(a – b) =

1 3 4   2 2 2 2

 cos(a – b) = 0 

sin(a  b)  1  0  1

23. If cos + cos + cos = 0 = sin + sin + sin, then (1) sin2 + sin2 + sin2 =

3 2

(3) cos2 + cos2 + cos2 =

(2) sin2 + sin2 + sin2 =

3 2

3 4

(4) cos2 + cos2 + cos2 = – 1

Sol. Answer (1, 3) Let a = cos + isin, b = cos + isin, c = cos + isin a+b+c=0 Also,

1 1 1   0 a b c

 ab + bc + ca = 0  a2 + b2 + c2 = 0  cos2 + cos2 + cos2 = 0 = sin2 + sin2 + sin2  2cos2 – 1 + 2cos2 – 1 + 2cos2 – 1 = 0  2cos2 + cos2 + cos2 =

3 2

1 – 2sin2 + 1 – 2sin + 1 – 2sin2 = 0  sin2 + sin2 + sin2 =

3 2

Trigonometric Equations 24. If 4cos – 3sec = 2tan, then  equals to

⎛ ⎞ (1) n  ( 1)n ⎜ ⎟ ⎝ 10 ⎠

n (2) n  ( 1)

 6

n (3) n  ( 1)

3 10

(4) n




127

Sol. Answer (1, 3) 4 cos – 3sec = 2tan 4cos2 – 3 = 2sin, cos  0 4sin2 + 2sin – 1 = 0

sin  

1  5 2 4

sin   1 

  n ( 1)n

5 1  5 , sin   4 4  3 or   n   ( 1)n 10 10

25. If   [–2, 2] and

⎛⎞ ⎛⎞ cos ⎜ ⎟  sin⎜ ⎟  2(cos36  sin18) , then the value of  is ⎝2⎠ ⎝2⎠ 7 6

(1)

(2)

 6

(3) 

5 6

(4) 

 6

Sol. Answer (1, 4) cos

   sin  2 (cos36º  sin18º ), [–2, 2] 2 2

⎛   ⎞ ⎛ 5  1⎞ ⎛ 5  1⎞ sin ⎜  ⎟  ⎜ ⎟⎜ ⎟ ⎝ 2 4 ⎠ ⎜⎝ 4 ⎟⎠ ⎜⎝ 4 ⎟⎠ 



1 2

     n   ( 1)n 2 4 6

    2n   ( 1)n .  3 2

  

 7 , 6 6

⎡ 7 ⎤ 26. If sin = a for exactly one value of  ⎢0, ⎥ , then the value of a is ⎣ 3⎦

(1)

3 2

(2) 1

(3) Zero

(4) –1


128



Sol. Answer (2, 4)

⎡ 7 ⎤ sin = a for exactly one value of   ⎢0, ⎥ ⎣ 3⎦ ∵ in [0, 2] sin takes every value twice except ± 1.

 a = ±1 27. If sin   3 cos   6y  y 2  11 ,   [0, 4], y  R holds for (1) No value of y and two values of 

(2) One value of y and two values of 

(3) Two values of y and one value of 

(4) Two pairs of (y, )

Sol. Answer (2, 4)

sin   3 cos   6y  y 2  11 ⎛ ⎛  ⎞⎞ 2 ⎜ sin ⎜   ⎟ ⎟  ( y 2  6 y  9)  9  11 3 ⎠⎠ ⎝ ⎝

⎞ ⎛ 2sin ⎜   ⎟  ( y  3)2  2 3 ⎝ ⎠ LHS [–2, 2], RHS  – 2  only solution exists if LHS = RHS = –2

⎞ ⎛ sin ⎜   ⎟  1 and (y – 3) = 0 3 ⎝ ⎠ ∵ [0, 4], y = 3  

 3 7  , 3 2 2

 Two values of  and one value of y. 28. The equation sin x  3 cos x  2 is satisfied by value of x which is equal to (1) x  n 

 , nI 12

(2) x  2n 

5 , nI 12

(3) x  n 

5 , nI 12

(4) x  2n 

 , nI 12

Sol. Answer (2, 4)

sin x  3 cos x  2 Or,

sin x 3 1  cos x  2 2 2

⎞  ⎛ Or, sin ⎜ x  ⎟  sin 3 4 ⎝ ⎠ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


 x


129

   n  ( 1)n 3 4

n = even ; x 

   2k   3 4

⎛  ⎞  x  2k   ⎜  ⎟ ⎝4 3⎠  ; k I 12

x  2k  

n = odd ; x 

   (2k  1)  3 4

   x  2k      4 3

29. If

 2k  

3   4 3

 2k  

5 ; k I 12

1 tan x   tan y and x  y  , then (x, y) is 1 tan x 6

⎛ 5  ⎞ (1) ⎜ , ⎟ ⎝ 24 24 ⎠

⎛ 7 11 ⎞ (2) ⎜  ,  ⎟ ⎝ 24 24 ⎠

⎛ 115 119 ⎞ , (3) ⎜  24 ⎟⎠ ⎝ 24

⎛ 5 13 ⎞ (4) ⎜ , ⎟ ⎝ 24 24 ⎠

Sol. Answer (1, 2, 3) 1  tan x   tan y , x  y  1  tan x 6

⎛ ⎞ tan ⎜  x ⎟  tan y 4 ⎝ ⎠ y  n 

  x and 4

x  y  n 

xy 

 x 

x–y 

 6

 4

 6

n 5 n    ,y  2 24 2 24


130



⎛ 5  ⎞ n  0; ( x, y )  ⎜ , ⎟ ⎝ 24 24 ⎠ ⎛ 7 11 ⎞ n  1; ( x, y )  ⎜  ,  24 ⎟⎠ ⎝ 24

⎛ 115 119 ⎞ n  10, ( x, y )  ⎜  , 24 ⎟⎠ ⎝ 24 30. The possible value of   [–, ] satisfying the equation 2(cos + cos2) + (1 + 2cos)sin2 = 2sin are (1) 

 2

(2) 

 3

(3)

 3

(4) 

Sol. Answer (1, 2, 3, 4)   [–, ] 2(cos + cos2) + (1 + 2cos)sin2 = 2sin or, 2(cos + cos2) + 2(cos + 2cos2)sin = 2sin or, 2(cos + cos2) + 2sin[cos + 2cos2 – 1] = 0 or, 2(cos + cos2) + 2sin(cos + cos2) = 0 or, 2(cos + cos2) (1 + sin) = 0 31. The number of all possible triplets (p, q, r) such that p + qcos2 + r sin2 = 0 for all , is (1) (k, –k, –2k)

⎛k k ⎞ (2) ⎜ ,  ,  k ⎟ ⎝2 2 ⎠

⎛ k k ⎞ (3) ⎜  , , k ⎟ ⎝ 2 2 ⎠

(4) (–k, k, 2k)

Sol. Answer (1, 2, 3, 4) p + q cos2 + r sin2 = 0

  R

or, p + q(1 – 2sin2) + rsin2 = 0 or, (r – 2q)sin2 + (q + p) = 0

  R

  R

 r – 2q = q + p = 0  r = 2q, q = –p 

r q p   2 1 1

Or,

p q r   1 1 2

32. A value of  satisfying sin sin(60 – ) sin(60 + ) =

(1)

4 3

1 sin3 is (  n) 4

(2) 

5 3

(3)

2 3

(4)

5 3




131

Sol. Answer (1, 2, 3, 4) 1 As, sin  sin(60   )sin(60   )  sin3 is an identity. So it is true for all the given options. 4

⎡  ⎤ 33. The co-ordinates of points of intersection of the curves y = cosx and y = sin3x, x  ⎢ , ⎥ are ⎣ 2 2⎦ ⎞ ⎛ (2) ⎜ , cos ⎟ 8⎠ ⎝8

⎛ 1 ⎞ (1) ⎜ , ⎟ ⎝4 2⎠

3 ⎞ ⎛ 3 (3) ⎜  , cos ⎟ 8 ⎠ ⎝ 8


y

cosx = sin3x

y = sin3x

1

⎛ ⎞ or, cos x  cos ⎜  3 x ⎟ ⎝2 ⎠

y = cosx

⎛ ⎞  x  2n  ⎜  3 x ⎟ 2 ⎝ ⎠  4 x  2n 

 x 

3 ⎞ ⎛ 3 (4) ⎜ , cos ⎟ 8 ⎠ ⎝ 8

 2

 3

 6

 6

 3

x

,  2 2

x

 2

  or,  2 x  2n  2 2

–1

2n    or, x  n  4 8 4

⎡  ⎤   3 So, the values of x in ⎢ , ⎥ is , and  4 8 8 ⎣ 2 2⎦ 34. A value of  satisfying 4cos2sin – 2sin2 = 3sin is (1)

9 10

(2)

 10

(3) 

13 10

(4) 

17 10

Sol. Answer (1, 2) sin[4cos2 – 2sin] = 3sin or, sin[4cos2 – 2sin – 3] = 0 or, sin[–4 sin2 – 2sin + 1] = 0 or, sin[4sin2 + 2sin – 1] = 0 

sin   0 ,

5 1  5 1 , 4 4

5 1 4

sin = 0

sin  

  = n

sin   sin



 10

⎛ ⎞   n  ( 1)n ⎜ ⎟ ⎝ 10 ⎠

sin   

5 1 4

sin    sin

3 10

n 1    n  ( 1)

3 10


132



35. The equation sinx + sin2x + 2sinx sin2x = 2cosx + cos2x is satisfied by values of x for which n (1) x  n  ( 1)

(3) x  2n 

 , nI 6

2 , nI 3

(2) x  2n 

2 , nI 3

(4) x  2n 

 , nI 2

Sol. Answer (1, 2, 3, 4) sinx + sin2x + 2sinx sin2x = 2cosx + cos2x or, (sinx + sin2x) – cos2x + (2sinx sin2x – 2cosx) = 0 or, sinx(1 + 2cosx) – cos2x + 2cosx(2sin2x – 1) = 0 or, sinx(1 + 2cosx) – cos2x – 2cosx(1 – 2sin2x) = 0 or, sinx(1 + 2cosx) – cos2x – 2cosx cos2x = 0 or, sinx(1 + 2cosx) – cos2x(1 + 2cosx) = 0 or, (1 + 2cosx) (sinx – cos2x) = 0  Either, cos x 



x  2n 

1 2

2 3

or, cos2x = sinx

or, 1 – 2sin2x = sinx or, 2sin2x + sinx – 1 = 0  sinx = –1,

1 2

 x  2n   , n  ( 1)n  , n  I 2 6 So, finally the general solution is x  2n 

  2 , 2n  , n  ( 1)n , n  I 3 2 6

36. The solution set of sin3 + cos2 = –2 is ⎧ ⎫ ⎧ n  ( 1)n 1 ⎬  ⎨ x x  n  (1) ⎨ x x  3 6 ⎩ ⎭ ⎩ ⎧ ⎫ n  ( 1)n 1 ⎬ n  I (3) ⎨ x x  3 6⎭ ⎩

⎫ ⎬ n I 2⎭

⎧ ⎫ ⎧ n  ( 1)n 1 ⎬  ⎨ x x  n  (2) ⎨ x x  3 6 ⎩ ⎭ ⎩

⎫ ⎬ n I 2⎭

⎧ ⎫ (4) ⎨ x x  n  ⎬ n  I 2⎭ ⎩

Sol. Answer (1, 2, 3, 4) sin3 + cos2 = –2 or, 3sin – 4sin3 + 1 – 2sin2 = –2 or, –4sin3 – 2sin2 + 3sin + 3 = 0 or, 4sin3 + 2sin2 – 3sin – 3 = 0 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



133

or, 4sin3 – 4sin2 + 6sin2 – 6sin + 3sin – 3 = 0 or, 4sin2(sin – 1) + 6sin(sin – 1) + 3(sin – 1) = 0  (sin – 1) [4sin2 + 6sin + 3] = 0  sin = 1 (∵ Another equation will not give real solutions)  sin = 1 is the only real solution   2n 



 ; n I 2

37. The general solution of the equation 2cos2 x  1  3·(2) sin

2

(2) n 

(1) n, n  I

 , nI 2

x

is

(3) n 

 , nI 2

(4) n 

 , nI 3

Sol. Answer (1, 2, 3) 2

212 sin

2

 1  3.2 sin

x 2

2  22 sin

x

22 sin2 x 2

2  22 sin 2sin

2

x

x



x

3 7 sin2 x 2

 3.2sin

x

y

 y2 – 3y + 2 = 0  (y – 1) (y – 2) = 0

y  2sin

2

x

 1 or 2sin

2

x

2

 sin2x = 0 or sin2x = 1 x = n, x  n 

38.

 2

(m  2)sin   (2m  1)cos   2m  1 , if

(1) tan  

3 4

(2) tan  

4 3

(3) tan  

2m 2

m 1

(4) tan  

2m m2  1

Sol. Answer (2, 3) (m + 2)sin + (2m – 1)cos = 2 m + 1  (m + 2)tan + (2m – 1) = (2m + 1)sec  (m + 2)2tan2 + (2m – 1)2 + 2(m + 2) (2m – 1)tan= (2m + 1)2.(1 + tan2)  (3m2 – 3) tan2 – 2(2m2 + 3m – 2) tan+ 8 m = 0 3(m2 – 1)tan2 – {4(m2 – 1) + 6m}tan+ 8 m = 0  (m2 – 1)tan. (3tan – 4) – 2m(3 tan– 4) = 0 tan  

4 2m or tan   2 3 m 1


134



Properties of Triangle a  2  3,  C  60  then in the triangle b (1) One angle is 105° (2) One angle is four times another angle

39. In a triangle ABC,

(3) One angle is 25°

(4) One angle is five times another angle

Sol. Answer (1, 2) In ABC Using sin formula

sin A a 2  3   sin B b 1

Apply components and dividends



sin A  sin B 3  3  sin A  sin B 1  3



AB AB cos 2 2  3 3 AB A  B 1 3 2cos sin 2 2

ab 3 3  a  b 1 3

2sin

 cot

AB 1 2

 A – B = 90° and A + B = 120° A = 105°  B = 15°  Option (1) and (2) are correct but option (3) & (4) are not correct. 40. In a triangle ABC (1) sinA.sinB.sinC =

 2R

2

(3) acosA + bcosB + ccosC =

abc 2R

2

(2) sinA.sinB.sinC =

r (sinA + sinB + sinC) 2R

(4) sinA.sinB.sinC =

R (sinA + sinB + sinC) 2r

Sol. Answer (1, 2) (1) sinA. sinB. sinC 





=

2 2 2   bc ac ab

83 2 2 2

a b c



1 ab sin C 2



abc 4R

83 16 2 R 2  2R 2

So option (1) is correct



(2) RHS

135

r (sin A  sin B  sin C ) 2R



r ⎛ 2 2 2 ⎞   2R ⎜⎝ bc ac ab ⎟⎠



r ⎛abc ⎞ .2 ⎜ ⎟ 2R ⎝ abc ⎠



r  2s  R abc



2 2 R.abc



2 2 R.4R




 2R 2

 LHS

So option 2 is correct (3) Since option (2) is correct option 4 is not correct Hence correct answers are (1, 2) ⎛C ⎞ 41. If cosA + cosB = 4sin2 ⎜ ⎟ , then ⎝2⎠

(2) cos C  1 

(1) 2sinB = sinA + sinC (3) cos A +cos B =

2r R

r R

(4) a, c, b are in G.P.

Sol. Answer (2, 3)

cos A  cos B  4 sin2  2cos  cos

C 2

AB AB C  4 sin2 cos 2 2 2

AB C  2 sin 2 2

 2cos

Multiply by 2cos

C both sides 2

C AB  2sin C cos 2 2

 sinA + sinB = 2sinC  a, b, c are in A.P. 

Option 1 is not correct


136



So we know that cosA + cosB + cosC = 1   4 sin2

r R

C r  cos C  1  2 R

(1  cos C )  cos C  1 

r R

r  cos C R  Option (2) is correct 1

Again cosA + cosB + cosC = 1  Put cosC = 1 

r R

r R

 cos A  cos B  1   cos A  cos B 

r r  1 R R

2r R

 Option (3) is correct As a, c, b are in A.P. and all three are not equal  a, c, b are not in G.P 42. In a triangle ABC, point D and E are taken on side BC such that BD = DE = EC. If angle ADE = angle AED = , then (1) tan = 3tanB

(2) 3tan = tanC

(3)

6 tan  tan2   9

 tan A

(4) B = C

A

Sol. Answer (1, 3, 4) Let BC = 6a  BD = DE = EC = 2a

h

AndDP = PE = a tan B 

h 3a

tan  

h a

 B

D

 P

E

C

6a

 tan = 3tanB  Option (1) is correct By figure it can be seen that B = C  Option (4) is correct And option (2) is incorrect Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


Now option (3); LHS 


6 tan  2

tan   9



6

h a

2

⎛h⎞ ⎜a⎟ 9 ⎝ ⎠



137

6ah 2

h  9a2

L.H.S. tanA = – tan(B + C) 

[tan B  tan C ] 1  tan B tan C

h h  tan B  tan C 6ah   3a2 3a  2 tan B tan C  1 h h  9a 2  1 9a 2

 Option (3) is correct 43. In a triangle, with usual notations, the length of the bisector of angle A is

2bc cos (1)

b+c

A 2

2bc sin (2)

b+c

A 2

A 2 2R(b + c )

abc cosec (3)

(4)

2 A cosec 2 b+c

Sol. Answer (1, 3, 4) Let AD = l Ar ABD + Ar ADC = ArABC 1 1 1 cl sin   bl sin   bc sin 2 2 2 2

1 1 l sin [b  c ]  bc 2sin  cos  2 2 l

2bc cos  2bc cos A / 2  bc bc

So option (1) is correct and option (2) is incorrect For option (3)



a 2bc cos A / 2 2abc cos A / 2 2abc cos A / 2   a bc (b  c ).2R sin A (b  c ).2R.2sin A / 2cos A / 2

=

abc cos ec A / 2 2R(b  c )

For option (4)

abc cosec A / 2 4R cosec A / 2 2cosecA / 2   2R(b  c ) 2R(b  c ) bc Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

138



44. An ordered triplet solution (x, y, z) with x, y, z  (0, 1) and satisfying x2 + y2 + z2 + 2xyz = 1 is  7 4 ⎞ ⎛ (1) ⎜ cos , cos , cos 6 18 9 ⎟⎠ ⎝

2   ⎞ ⎛ (2) ⎜ cos , cos , cos ⎟ 5 3 10 ⎠ ⎝

7  ⎞ ⎛ (3) ⎜ cos , cos , cos ⎟ 12 4 6 ⎝ ⎠

4 17 ⎞  ⎛ (4) ⎜ cos , cos , cos 12 9 36 ⎟⎠ ⎝

Sol. Answer (1, 4) Let A  cos1 x, B  cos1 y , C  cos1 z, where A + B + C =  A+B+C= cos–1x + cos–1y + cos–1z =  cos–1x + cos–1y =  – cos–1z

cos–1[ xy  1  x 2 1  y 2 ]  cos1(3) xy  z  1  x 2 1  y 2

(xy + z2) = (1 – x2) (1 – y2) x2y2 + z2 + 2xyz = 1 – x2 – y2 + x2y2 x2y2 + z2 + 2xyz = 1 which is the given identity  A+B+C= which is satisfied by options (1) and (4) and A, B, C should be acute.

45. In a ABC, acosB + bcosC + ccosA =

abc , then 2

(1) Triangle in isosceles

(2) Triangle may be equilateral

(3) sin(A – B) + sin(B – C) + sin(C – A) =

3 2

⎛ A  B ⎞ ⎛ B C ⎞ ⎛C  A ⎞ (4) 4 sin ⎜ ⎟ sin ⎜ ⎟ sin ⎜ ⎟ =1 ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠

Sol. Answer (1, 2) In ABC a cosB + b cos C + c cosA =

abc 2

2RsinAcosB + 2RsinBcosC + 2RsinCcosA =

2R sin A  2R sin B  2R sin C 2

sin(A + B) + sin(A – B) + sin(B + C) + sin(B – C) + sin(C + A) + sin(C – A) = sinA + sinB + sinC  sin(B + C) + sin(C – A) + sin(A + B) = 0 

4 sin

 sin

BA A C C B 0 sin sin 2 2 2

BA A C C B  0 or sin  0 or sin 0 2 2 2




139

 B = A or A = C or C = B Option (1) is correct Option (2) is correct

(If A = B = C)

Option (3) is not correct Option (4) is not correct 46. In a triangle ABC with fixed base BC, the vertex A moves such that cos B  cos C  4 sin2

A 2

If a, b and c denote the lengths of the sides of the triangle opposite to the angles A, B and C respectively, then [IIT-JEE 2009] (1) b + c = 4a

(2) b + c = 2a

(3) Locus of point A is an ellipse

(4) Locus of point A is a pair of straight lines

Sol. Answer (2, 3) 2 Given cos B  cos C  4 sin

A 2

A ⎛B C ⎞ ⎛ B C ⎞ .cos ⎜  4 sin2  2cos ⎜ ⎟ ⎟ 2 ⎝ 2 ⎠ ⎝ 2 ⎠ A ⎛ A⎞ ⎛ B C ⎞  4 sin2  2sin ⎜ ⎟ .cos ⎜ ⎟ 2 ⎝2⎠ ⎝ 2 ⎠ A ⎛ B C ⎞ A  cos ⎜ ⎟  2sin 2 , as sin  0 2 2 ⎝ ⎠ ⎛ B C ⎞ cos ⎜ ⎟ ⎝ 2 ⎠ 2  ⎛B C ⎞ cos ⎜ ⎟ ⎝ 2 ⎠ B C 1  tan .tan 2 2 2  B C 1  tan .tan 2 2



1

(s  a )(s  c ) (s  a )(s  b ) . s (s  b ) s (s  c )

1

(s  a )(s  c ) (s  a )(s  b ) = 2 . s(s  b ) s( s  c )

(s  a ) s 2  s a 1 s 1


140




2s  a 2 a



bc 2 a


b  c  2a and also b + c = 2a  AC + AB = 2BC  AC + AB > BC, which shows that the locus of point A is an ellipse.

47. If

sin4 x cos4 x 1   , then 2 3 5

2 (1) tan x 

2 3

[IIT-JEE 2009]

sin8 x cos8 x 1   8 27 125

(2)

2 (3) tan x 

1 3

(4)

sin8 x cos8 x 2   8 27 125

Sol. Answer (1, 2) We have,

sin4 x cos4 x 1   2 3 5  3 sin4x + 2cos4x =

6 5

 3(1 – cos2x)2 + 2cos4x =

6 5

 3(1 + cos4x – 2 cos2x) + 2cos4x =

 3 + 3cos4x – 6cos2x + 2cos4x =

4 2  5 cos x  6 cos x  3 

4 2  5 cos x  6 cos x 

6 5

6 5

6 0 5

9 0 5

 25cos4x – 30cos2x + 9 = 0  (5cos2x – 3)2 = 0 2  cos x 

2  sec x 

3 5

5 3



tan2 x  sec 2 x  1 


141

5 2 1 3 3

Hence option (1) is true sin2 x  1  cos2 x  1 

3 2  5 5 4

4

⎛2⎞ ⎛3⎞ sin x cos x ⎜ ⎟ ⎜ ⎟ 5  Now, = ⎝ ⎠  ⎝ 5 ⎠  2  3  5  1 option (2) 8 27 8 27 5 4 5 4 5 4 125 8

48. For 0   

(1)

8

 , the solution(s) of 2

 4

(2)

6

⎛

∑ cosec ⎜⎝  

m 1

 6

(m  1) ⎞ m ⎞ ⎛ ⎟ cosec ⎜   4 ⎟  4 2 is(are) 4 ⎠ ⎝ ⎠ (3)

 12

(4)

[IIT-JEE 2009] 5 12

Sol. Answer (3, 4) We have m   ⎞ m ⎞ ⎛ ⎛ Tm  cosec ⎜   cosec ⎜   ⎟ 4 ⎠ 4 ⎟⎠ ⎝ ⎝



1 m  ⎞ ⎛ m ⎞ ⎛  ⎟ sin ⎜   sin ⎜   4 4⎠ ⎝ 4 ⎟⎠ ⎝

⎡ ⎛⎛ m ⎞ ⎛ m  ⎞ ⎞ ⎤  ⎜   ⎟⎟ ⎥ ⎢ sin ⎜ ⎜   ⎟ 4 ⎠ ⎝ 4 4 ⎠⎠ ⎥ 1 ⎢ ⎝⎝  ⎢ m ⎞ ⎛ m  ⎞ ⎥ sin ⎢ sin ⎜⎛   sin ⎜    ⎟⎥ ⎟ 4⎢ 4 ⎠ ⎝ 4 4 ⎠ ⎥⎦ ⎝ ⎣

⎡ ⎛ m ⎞ m  ⎞ m ⎞ ⎛ m  ⎞ ⎤ ⎛ ⎛ ⎢ sin ⎜   4 ⎟ cos ⎜   4  4 ⎟  cos ⎜   4 ⎟ sin ⎜   4  4 ⎟ ⎥ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎥  2⎢ ⎝ ⎢ ⎥ m ⎞ ⎛ m  ⎞ ⎛ sin ⎜   ⎟ sin ⎜   4  4 ⎟ ⎢ ⎥ 4 ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ ⎡ ⎛ m  ⎞ m ⎞ ⎤ ⎛  2 ⎢cot ⎜    ⎟  cot ⎜   ⎥ 4 4⎠ 4 ⎟⎠ ⎦ ⎝ ⎣ ⎝ ⎛  ⎞⎞ ⎛  T1  2 ⎜ cot()  cot ⎜   ⎟ ⎟ 4 ⎝ ⎠⎠ ⎝ ⎛ 2 ⎞ ⎞ ⎞ ⎛ ⎛ T2  2 ⎜ cot ⎜   ⎟  cot ⎜   ⎟ 4⎠ 4 ⎟⎠ ⎠ ⎝ ⎝ ⎝ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

142



⎛ 5 ⎞ 6 ⎞ ⎞ ⎛ ⎛ T6  2 ⎜ cot ⎜    cot ⎜   ⎟ ⎟ 4 ⎠ 4 ⎟⎠ ⎠ ⎝ ⎝ ⎝

Sum T1  T2  T3  T4  T5  T6  4 2 

⎡ 6 ⎞ ⎤ ⎛ 2 ⎢cot   cot ⎜   ⎥4 2 4 ⎟⎠ ⎦ ⎝ ⎣

cot + tan = 4 

sin2   cos2  4 sin  cos 

 sin  cos   2sin  cos  

1 4

1 2

sin 2 

1 2

 2 

 5 , 6 6

 

 5 , 12 12

 and let a, b and c denote the lengths of the sides opposite to A, B 6 and C respectively. The value(s) of x for which a = x 2 + x + 1, b = x 2 – 1 and c = 2x + 1 is/are [IIT-JEE 2010]

49. Let ABC be a triangle such that ACB 



(1)  2  3



(2) 1  3

(3) 2  3

(4) 4 3

Sol. Answer (2)

cos C 



2 2 2 b a c 2ab

    2



2

x 2  1  x 2  x  1   2 x  1 3  2 2 x2  1 x2  x  1

2



 ( x  2)( x  1)x( x  1)  ( x 2  1)2  3( x 2  x  1)( x 2  1) 

C

3( x 2  x  1)  x 2  2 x  x 2  1

 6

 (2  3)x 2  (2  3)x  ( 3  1)  0  x  (2  3) & x  1  3

A

B

But x  (2  3) shows that c is negative, which is not possible in a triangle. Hence x  1  3 is only possible value of x. ⇒ x  1  3 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



143

50. Let ,   [0, 2] be such that

 ⎞ ⎛ 2cos (1  sin )  sin2  ⎜ tan  cot ⎟ cos   1 , 2 2⎠ ⎝ tan(2  )  0 and 1  sin   

3 . 2

Then  cannot satisfy (1) 0   

 2

[IIT-JEE 2012]  4  2 3

(2)

(3)

4 3  3 2

(4)

3    2 2


⎛ ⎞ ⎜ ⎟ 1 2cos  1  sin    sin2  ⎜ ⎟ cos   1   ⎜⎜ sin ·cos ⎟⎟ ⎝ 2 2⎠  2cos  1  sin    2sin  ·cos   1  2cos + 1 = 2sin( + )

…(i)

Now, tan   0 and  1  sin  

 3 gives 2

⎛ 3 5 ⎞ ⎜ , ⎟ ⎝ 2 3 ⎠ Using (i) ; sin       cos   

1 2

1  sin(  )  1 2

i.e.,

 5 13 17     or, 6 6 6 6



13 17    6 6



2 7  3 6

So, options (1), (3) and (4) are correct.

51. In a triangle PQR, P is the largest angle and cos P =

1 . Further the incircle of the triangle touches the sides 3

PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are) [JEE(Advanced)-2013] (1) 16

(2) 18

(3) 24

(4) 22


144



Sol. Answer (2, 4) QR is largest side PM = PN = 2k RM = RL = 2k + 4 QL = QN = 2k + 2  QR = 4k + 6 RP = 4k + 4 QP = 4k + 2

cosP =

1 = 3

 PQ 2   PR 2  QR 2 2  PQ   PR 

R

 4 k  2  2   4 k  4  2   4k  6  2 2  4 k  2   4k  4 

1  2k  1  4  k  1   2k  3   3 4  2k  1  k  1 2

2

+ 2k 2k +

4

4

L

+ 2k

2Q

2k + 2 N M

2k 2k

P

2

4  k  1   4k  4   2  1 = 4  2 k  1  k  1 3 2

1 = 3

 k  12 – 2  k  1  k  1  2k  1

(k + 1) (2k + 1) = 3(k – 1) (k + 1) k = – 1 or 2k + 1 = 3k – 3 4 = k So, PQ = 18, QR = 22, RP = 20

SECTION - C Linked Comprehension Type Questions Comprehension - I If Pn  cosn   sinn , 1.

6P10  15P8  10P6 is equal to

(1) –1 2.

(2) 0

(3) 1

(4) 8

(3) 4

(4)

Maximum value of P1000 will be (1) 2

(2) 1

1 2




⎡ ⎤   ⎢0, ⎥ , n  ( , 2) , then minimum of Pn will be ⎣ 2⎦ 1 (1) 1 (2) 2 Solution of Comprehension - I

145

3.

1.

Answer (3)

2.

Answer (2)

3.

Answer (1)

(3)

2

(4)

1 2

Pn = cosn + sinn Pn = cosn + sinn Pn – 2 = cosn – 2 + sinn – 2 Pn – Pn – 2 = (cosn  + sinn) – (cosn – 2  + sinn – 2) = cosn – 2 (– sin2) + sinn – 2  (–cos 2) = – sin2 cos2 (Pn – 4) 1.

 6P10 – 15P8 + 10P6 = 6P10 – 6P8 – 9P8 + 9P6 + P6 – P4 + P4 – P2 + P2 = 6(–sin2cos2  P6) – 9(–sin2 cos2.P4) + (–sin2cos2P2) + (–sin2cos2P0) + 1 = sin2cos2  [–6P6 + 9P4 – P2) – 2] + 1 = sin2cos2  [–6(P6 – P4) + 3(P4 – P2) – 2P2 – 2] + 1 = sin2cos2  [–6  (–sin2 cos2P2) + 3(sin2 cos2P0) + 2 – 2] + 1 = sin4cos4[– 6P2 + 3P0] + 1 = sin4cos4[– 6 + 6] + 1 = 1

2.

Maximum value of P1000 = ? cos1000 + sin1000 = cos2 cos1998 + sin2 sin1948 cot2 + sin2 = 1  (P1000)max = 1

3.

⎡ ⎤ Pn = cosn + sinn,   ⎢0, 2 ⎥ , 2  ( , 2) ⎣ ⎦ Minimum value of Pn = ?

sin2  (sin)n and cos2 (cos)4 

⎛ ⎞  n  (– , 2) &   ⎜ 0, ⎟ ⎝ 2⎠

⎛ ⎞  n  (– , 2) &   ⎜ 0, ⎟ ⎝ 2⎠

sinn   cosn   1

 (Pn)min = 1 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

146



Comprehension-II From algebra we know that if ax2 + bx + c = 0; a( 0), b, c  R has roots  and  then     

c b and   . a a

Trigonometric functions sin and cos; tan and sec; cosec and cot obey sin2 + cos2 = 1. A linear relation in sin and cos, sec and tan or cosec and cot can be transformed into a quadratic equation in, say, sin, tan or cot respectively. And then one can apply sum and product of roots to find the desired result. Let acos + bsin = c have two roots 1 and 2, 1  2. 1.

The value of cos(1 + 2) is (a and b not being simultaneously zero)

(1)

a2  b2

(2)

a2  b2

b2  a2

(3)

a2  b2

a2  c 2

(4)

a2  b2

c 2  a2 a2  b2

Sol. Answer (1) acos + bsin = c (acos – c)2 = ab2sin2 = b2 – b2cos2 a2cos2 – 2accos + c2 – b2 + b2cos2 = 0 (a2 + b2)cos2 – 2accos + (c2 – b2) = 0

cos 1  cos 2 

2ac a2  b

, cos 1 cos 2  2

c 2  b2 a2  b2

a2cos2 = c2 + b2sin2 – 2bcsin a2 – b2sin2 – b2sin2 + 2bcsin – c2 = 0  (a2 + b2) – 2bcsin + (c2 – a2) = 0  sin1 + sin2 =

sin1 sin2 =

2bc 2

a  b2

c 2  a2 a2  b2

 cos(1 + 2) = cos1cos2 – sin1sin2 



2.

c 2  b2 a2  b2



c 2  a2 a2  b2

a2  b2 a2  b2

The value of cos(1 – 2) is (a and b not being simultaneously zero) (1) 1 

2c 2 a2  b2

(2)

2c 2 a2  b2

1

(3) 1 

2c 2 2

a b

2

(4)

2c 2 2

a  b2

1




147

Sol. Answer (4) cos(1 – 2) = cos1cos2 + sin1sin 2  

3.

c 2  b2  c 2  a2 a2  b2 2c 2 a2  b2

1

The values of tan1tan2 is (given |b| |c|) (1)

a2  c 2

(2)

a2  b2

a2  c 2 a2  b2

(3)

a2  c 2 c 2  b2

(4)

c 2  a2 c 2  b2

Sol. Answer (4)

tan 1 tan 2 



c 2  a2 c 2  b2



a2  b2 c 2  b2

c 2  a2 c 2  b2

Comprehension-III In reducing a given trigonometric equation to the standard form (sinx = sin) or (cosx = cos) etc. we apply several trigonometric or Algebraic transformation. As a result of which final form so obtained may not be equivalent to the original equation resulting either in loss of solutions or appearance of extraneous solutions. 1.

The solution set of the equation

5  2sin x  6sin x  1 is (1) n + (–1)n where sin  

(2) x  n  ( 1)n

 2 or x  n  ( 1)n , n  I 6 9

 , nI 6

(3) x = n + (–1)n where sin  

2 9

(4) Null set Sol. Answer (2) 5  2 sin x  6 sin x  1

5 – 2sinx = 36sin2x – 12sinx + 1 36sin2x – 10sinx – 4 = 0 18sin2x – 5sinx – 2 = 0 18sin2x – 9sinx + 4sinx – 2 = 0 (9sinx + 2) (2sinx – 1) = 0 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

148


Putting sin x 

sin x 

1 2 , sin x   2 9

1 in given equation 2

LHS  5  2 

RHS = 6 

1 2 2

1 1 = 2 2

Putting sin x  

LHS =


5

RHS = 6  

2 9

4 7  9 3 2 1 0 9

No solution  sin x 

1 is a solution 2

n  x  n   ( 1)

2.

 ;nI 6

The equation 2cot2x – 3cot3x = tan2x has

⎛ ⎞ (1) Two solutions in ⎜ 0, ⎟ ⎝ 3⎠

⎛ ⎞ (2) One solution in ⎜ 0, ⎟ ⎝ 3⎠

(3) No solution in (– , )

(4) Three solution in (0, )

Sol. Answer (3) 2cot2x – 3cot3x = tan2x 2(cot2x – cot3x) = tan2x + cot3x  2(cos 2 x sin3 x  cos3 x sin2x )  sin2 x.sin3 x  cos 2 x.cos3 x sin2x .sin3 x cos 2 x.sin3 x 

2 sin x cos x  sin 2 x.sin3 x cos 2 x.sin3 x

2tanx = tan2x 1 – tan2x = 1 tan2x = 0 But tanx  0 as given equation involves terms of cotangent.  No solution Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


3.

149

13  18tan x  6tan x  3 is

The solution of (1) tan x  


1 6

1⎞ ⎛ 2 ⎞ ⎛ (3) tan x  ⎜ ,  ⎟  ⎜ ,  ⎟ 6 ⎝ ⎠ ⎝3 ⎠

(2) tanx =

2 3

(4) tan x 

3 4

Sol. Answer (2)

13  18 tan x  6 tan x  3  13 – 18tanx = 36tan2x – 36tanx + 9  36tan2x – 18tanx – 4 = 0  18tan2x – 9 tanx – 2 = 0 18 tan2x – 12tanx + 3tanx – 2 = 0 (6tanx + 1) (3tanx – 2) = 0 tan x 

1 2 , tan x  6 3

Putting tan x   L.H.S. =

1 in given equation, 6

13  3  4 R.H.S. = –1 – 3 = –4

 No solution Putting tan x  L.H.S. =

2 in given equation 3

13  12  1

R.H.S. = 4   tan x 

2 is a solution. 3

Comprehension-IV There are trigonometrical equations that are nonstandard in aspect in that the equation may contain dissimilar terms, say algebraic as well as trigonometric expressions, or there may be only one equation in more than one unknowns. In such cases we make use of the extreme values of trigonometric functions or algebraic functions. In particular |sinx|  1, |cosx|  1, and maxima/minima of quadratic expressions are widely made use of.

1.

⎛ x ⎞ The equation sin⎜ ⎟  x 2  6 x  10 holds for ⎝ 6 ⎠

(1) Infinitely many values of x

(2) Finitely many values of x

(3) Just one value of x

(4) No value of x


150



Sol. Answer (3) ⎛ x ⎞ sin ⎜ ⎟ = (x – 3)2 + 1 1,  x  R ⎝ 6⎠ ⎛ x ⎞ We know sin ⎜⎝ ⎟⎠  1 6

Equality holds at x = 3. Hence the given equation has only one solution.

2.

4 The equation 8 cos

x x 1 sin2  x 2  , x  (0, 4] holds for 2 2 x2

(1) No value of x (2) Exactly two values of x, both greater than  (3) Exactly two values of x, one smaller than  and the other greater than  (4) Just one value of x Sol. Answer (1)

2cos2  cos

x 1 sin2 x  x 2  2  2  x  R, except x = 0 2 x 2

x sin2 x  1 (which is not possible) 2

2 Because 0  cos

2 Hence cos

x sin2 x  1 2

2 But if cos

x sin2 x  1 2

2 Then cos

x  1 and sin2 x  1 2



3.

x  1 and 0  sin2 x  1 2

x   n and x  k   n, k  Z , which is not possible simultaneously. Hence no solution exits. 2 2

The equation cosec

x y z   cosec  cosec  6, where 0  x, y , z  and x + y + z = , have 2 2 2 2

(1) Three ordered triplet (x, y, z) solutions

(2) Two ordered triplet (x, y, z) solutions

(3) Just one ordered triplet (x, y, z) solution

(4) No ordered triplet (x, y, z) solution

Sol. Answer (3) cosec

x y z  cosec  cosec = 2 + 2 + 2 2 2 2

 x=y=z=

 3




151

Comprehension-V Recall that sinx + cosx = u (say) and sinxcosx = v (say) are connected by (sinx + cosx)2 = sin2x + cos2x + 2sinxcosx  u2 = 1 + 2v  v=

u2 1 2

It follows that any rational integral function of sinx + cosx, and sinxcosx i.e., R(sinx + cosx, sinxcosx), or in our ⎛ u 2  1⎞ ⎟ . Thus to solve an equation of the form R(u, v) = 0, we form notation R(u, v) can be transformed to R ⎜⎜ u, 2 ⎟⎠ ⎝

a polynomial equation in u and then look for solutions. 1.

The solution set of sin x  cos x  2 2 sin x cos x  0 is completely described by (1) x  2n 

 5 11 , 2n  , 2n  , nZ 4 12 12

(2) x  2n 

  7 , 2n  , 2n  , nZ 4 12 12

(3) x  2n 

  7 , 2n  , 2n  , nZ 4 12 12

(4) x  2n 

  7 , 2n  , 2n  , nZ 4 12 12

Sol. Answer (1) sinx + cosx = 2 2 sinx cosx

⎛ u 2  1⎞ u  2 2 v  2 2  ⎜⎝ 2 ⎟⎠

 u

1 1 8 2 2

 2, 

1 2



2u 2  u  2  0

 sin x + cos x =

1 2,  2

⎞ 1 ⎛ ⎛ ⎞  sin ⎜⎝ x  ⎟⎠    sin ⎜⎝  ⎟⎠ 4 2 6  x + sin

   n   ( 1)n 1. 4 6

⎞  ⎛ and sin ⎜⎝ x  ⎟⎠  1  sin 4 2  x

   n   ( 1)n 4 2

 x  n 

    ( 1)n = n   for n is even or odd. 4 2 4


152 2.



The complete solution of the equation sin2x – 12(sinx – cosx) + 12 = 0 is given by (1) x  2n 

 , (2n  1), n  Z 2

(2) x  n 

 , (2n  1), n  Z 2

(3) x  2n 

 , (2n  1), n  Z 2

(4) x  n 

 , (2n  1), n  Z 2

Sol. Answer (3) Let u = cosx – sinx, v = sinx cosx =

1 u2 2

Thus the given equation reduces to 2v + 12u + 12 =0  1 – u2 + 12u + 12 = 0  u2 – 12u – 13 = 0  (u + 1) (u – 13) = 0  cosx – sinx = –1 as u  13

⎛  cos ⎜⎝ x   x

⎞ 3 ⎟⎠  cos 4 4

 3  2n   4 4

 x  2n   3.

 , (2n  1), n  Z (by taking positive and negative respectively) 2

The number of solutions of the equation sin + cos = 1 + sincos in the interval [0, 4] is (1) Four

(2) Six

(3) Eight

(4) Five

Sol. Answer (4) Let y  sin x  cos x Then the given equation can be reduced to y  1

y2 1 2

 y 2  2 y  1  0  y=1  sin x + cos x = 1

 x

⎞ 1 ⎛  sin ⎜⎝ x  ⎟⎠  4 2

       = ,   , 2  ,3  , 4  , 5  4 4 4 4 4 4 4

 x  0,

 5 9 , 2, , 4, 2 2 2

Hence in the given interval x  0,

 5 , 2, , 4 five solution exists. 2 2




153

Comprehension-VI Consider a triangle ABC, and that AA, BB, CC be the perpendicular from A, B and C upon the sides opposite to them. These three perpendiculars meet in H, called the orthocentre of the triangle. The triangle ABC formed by the feet of the perpendicular is called the pedal triangle of ABC. (Assume A, B, C  90°)

A C

B 1.

H

B

C

A

Suppose the triangle ABC have angles 60°, 70° and 50°. Then the pedal triangle ABC have angles given by (1) 80°, 60°, 40°

(2) 120°, 40°, 20°

(3) 30°, 65°, 85°

(4) 45°, 55°, 80°

Sol. Answer (1) A =  – 2A = –120 = 60° B =  – 2B = –140 = 40° C =  – 2C = –100 = 80° 2.

Suppose a triangle ABC has its sides 13, 14 and 15 cm. Then the circumradius of the pedal triangle is (in cm)

(1)

65 24

(2)

65 8

(3)

26 3

(4)

65 16

Sol. Answer (4) Given a = 13, b = 14, c = 15 Let R is circum radius of ABC, then

R

abc 4

Now, 2s = 13 + 14 + 15 = 42 s = 21   s(s  a )(s  b )(s  c ) =

21 8  7  6  84

R

13  14  15 65  4  21 4 8

Now, circum radius of pedal triangle =

R 65  2 16


154 3.



Suppose ABC is an acute angled triangle, then the area of the pedal triangle is (R being the circum radius of triangle ABC) (1)

R2 sin2A sin2B sin2C 2

(2) R2sin2A sin2B sin2C

(3)

R2 sin A sin B sin C 2

(4) R2sinA sinB sinC

Sol. Answer (1) Area of DEF 

DE  EF  FD 4  R

where R’ is the circumradius of DEF =

(a cos A)(b cos B )(c cos C ) ⎛ R⎞ 4⎜ ⎟ ⎝ 2⎠

=

(2R sin A cos A)(2R sin B cos B )(2R sin C  cos C ) 2R

=

R3 sin2 A  sin2B  sin2C 2R

=

1 2 R sin 2 A  sin 2B  sin 2C 2

Comprehension-VII Let us consider a triangle ABC having BC = 5 cm, CA = 4 cm, AB = 3 cm D, E are points on BC such that BD = DE = EC, CAE = , then 1.

AE2 is equal to (1)

73 3

(2)

73 5

(3)

73 7

(4)

73 9

Sol. Answer (4) Given BD = DE = EC = x(let) And ACE =  In ACE, by using cosine rule

cos  

cos  

5 AE 2  16  CE 2 and CE  3 2 AE  4

AE 2  16  2  AE  4

25 9

…(i)




155

In AEB, by using sine rule

sin(90 – ) sinB  BE AE cos  

BE 10 4  sin B  AE 3.AE 5

…(ii)

From equation (i) and (ii)

10 4   3  AE 5

AE 2  16  2  AE  4

25 9

64 119  AE 2  3 9 2  AE 

2.

73 9

tan is equal to (1)

3 4

(2)

1 2

(3)

3 8

(4)

5 8

(2)

52 9

(3)

52 7

(4) 52

Sol. Answer (3) From equation (ii) cos  

cos  

8 3  AE

8  ⎛ 73 ⎞ 3⎜ ⎟ ⎝ 3 ⎠

 tan   3.

8 73

3 8

AD2 is equal to (1)

52 3

Sol. Answer (2) In ABD, by using cosine rule

cos B 

3  5

9

BA2  BD 2  AD 2 2BA  BD 25  AD 2 9 5 23 3


156


6  9

25  AD 2 9

2  AD 

AD 2 


25 3 9

52 9

Comprehension-VIII Given a right angled triangle ABC with right angled at C. C = 90° and a, b, c are length of corresponding sides. (b > a). B

 c

a2 + b2 = c2

( A  B )  90 1.

A

a

b

C

cos(A – B) will be equal to (1) cos2B

(2) sin2A

(3) cos2A

(4) sinB

Sol. Answer (2)  cos(90° – B – B) = cos(90° – 2B) = sin2B Also cos(A – (90° – A)) = cos(2A – 90°) = sin(2A) 2.

sin2A will be equal to

2bc

(1)

(2)

a2

b2  a2 c2

(3)

2ab c2

(4)

b2  c 2 2bc

(4)

a c b

Sol. Answer (3)  sin2A = 2sinA cosA

 2



a b  c c

2ab c2

c2 – b2 = a2 3.

tan

(1)

A will be equal to 2

2ab 2

a b

2

(2)

2ab 2

b a

2

(3)

c b a




157

Sol. Answer (3)

A 1  cos A tan2   2 1  cos A



(c  b )2 c 2  b2

⇒ tan



b c  c b b cb 1 c

1

(c  b )2 a2

A cb  2 a

SECTION - D Assertion-Reason Type Questions 1.

STATEMENT-1 : Minimum value of sinx + cosecx is 2 for all x  (0, ). and STATEMENT-2 : Min (sinx + cosecx) = 2,  x  R.

Sol. Answer (3) Statement (2): Min(sinx + cosecx) = 2,  x R False ∵ (sinx + cosecx)min = 2 If sinx, cosecx are positive Statemenet (1) minimum value of (sinx + cosecx) = 2 when x R(0, ) is true 2.

STATEMENT-1 : If cos(A + B)sin(C + D) = cos(A – B)sin(C – D), then the value of cotAcotBcotC is cotD. and STATEMENT-2 : If cos(A + B)sin(C + D) = cos(A – B)sin(C – D), then the value of tanAtanBtanC is tanD.

Sol. Answer (1) From Statement-2

cos( A  B ) sin(C  D )  cos( A  B ) sin(C  D ) 1 tan A tan B tanC  tan D  1 tan A tan B tanC  tan D  tanA · tanB · tanC = tanD  cotA · cotB · cotC = cotD

3.

STATEMENT-1 : If cos  =

1 13 and cos  = where  and  both are acute angles, then the value of  –  is 7 14

 . 3

and  1 STATEMENT-2 : cos = 3 2 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

158



Sol. Answer (4) cos( – ) = cos · cos + sin · sin

4.



1 13 48 27 ·  · 7 14 7 14



13  36 49  91 91

STATEMENT-1 : The value of k for which (cos – sin)2 + ksincos – 1 = 0 is an identity is –2. and STATEMENT-2 : An identity in  is satisfied by all real values of .

Sol. Answer (4) sin2 + cos2 – 2sin · cos + ksin · cos – 1 (k – 2)sin · cos = 0  k = 2 Statement-1 is false, Statement-2 is true.

5.

STATEMENT-1 : If is acute and 1 + cos = k, then sin

 is 2

2k . 2

and 2 STATEMENT-2 : 2sin

x = 1  cos x. 2

Sol. Answer (1) 1 + cos = k 1 1 2sin2

 k 2

2k   sin2 2 2

 2k  sin  2 2

Statement-1 is true, Statement-2 is true and Statement-2 is correct explanation of (1).

6.

STATEMENT-1 : The general solution of 2

sin x

cos x

2

1

2

1 2

is n 

 . 4

and STATEMENT-2 : A.M.  G.M. Sol. Answer (1) Statement-1

2

sin x

cos x

2

1

2

1 2




159

sin x  2cos x  2 (2sin x 2cos x ) L.H.S. = 2

L.H.S.  2

sin x  cos x 1 2 1

1

 L.H.S.  2

2

1

∵ R.H.S. = 2

1 2

Equality holds good if 2sin x  2cos x  sinx = cosx x  n 



 4

 Both statements are correct and statement-2 explains statement-1 correctly. 7.

STATEMENT-1 : The equation 3cosx + 4sinx = 6 has no solution. and STATEMENT-2 : Due to periodic nature of sine and cosine functions, the equation 3cosx + 4sinx = 6 has infinitely many solutions.

Sol. Answer (3) –5  3 cos x + 4sinx  5  Given equation has no solution & statement 2 is false. n

8.

STATEMENT-1 : If

∑ sin( xr )  n , then

r 1

n

∑ cot( xr )  n .

r 1

and STATEMENT-2 : The number of solutions of the equation cosx = x is 1. Sol. Answer (2) Statement (2) cosx = x Only one solution true Statement-1 n

If

∑

r 1

sinxr = n, then cotxr = n

sinx1 + sinx1 + …. + sinxn = n  sin x1  sin x2  .....  sin xn  1 

x1  x2  .....  xn 

 2

 cot x1  cot x2  ....  cot xn  n  1 + 1 + …. + 1 = n is also true  Both statement are correct but statement -2 is not the correct explanation of statement -1. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

160 9.



STATEMENT-1 : The number of real solution of the equation sinx = 4x + 4–x is zero. and STATEMENT-2 : | sinx |  1 x  R.

Sol. Answer (1) Statement (2) : | sin x |  |  x  R is true. x x Statement (1) : sin x  4  4 .

L.H.S.  [–1, 1]

(as from statement-2)

R.H.S.  No solution  Both statements are correct and statement -2 is the correct explanation of statement -1.

10. STATEMENT-1 : The general solution of tan 5 = cot 2 is  

n  , nZ .  7 14

and STATEMENT-2 : The equation cos = k has exactly two solutions in [0, 2] for all k, –1  k  1. Sol. Answer (3) tan5 = cot2

y

It is clear from the graph 5 = n 

Hence generalizing  =

tan 5

 – 2 , n  Z 2

n   and 7 14

The equation cos x = k has only one solution in [0, 2  ]

 10

0

 10

for k = {–1, 1} and has three solutions for k = 0.

 5

3 10

 2

x

cot 2

 Statement-2 is false. 11. STATEMENT-1 : sinx = siny  x = y. and STATEMENT-2 : sinx = siny has infinitely many solutions for real values of x and y. Sol. Answer (4) Statement (2) is correct thus clearly Statement (1) is wrong sin x = sin y  (x = y,  – y……) ⎛ ⎞ 12. STATEMENT-1 : f(x) = logcosxsinx is well defined in ⎜ 0, ⎟. ⎝ 2⎠ and

⎛ ⎞ STATEMENT-2 : sinx and cosx are positive in ⎜ 0, ⎟. ⎝ 2⎠ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



161

Sol. Answer (2) Statement (2) is correct as both sinx and cosx are positive in I quadrant Statement (1) f ( x )  logcos x (sin x )

=

lnsin x lncos x

Is defined if cosx > 0 and cosx 1 ⎛ ⎞  It is defined in ⎜⎝ 0, 2 ⎟⎠ (as given)

 Statement (1) is true but (2) is not correct explanation for (1) 13. STATEMENT-1 : If in ABC, 3bc = (a – b+c)(a + b – c) then A = 120°. and STATEMENT-2 : cos120° = 

1 2

Sol. Answer (2) 3bc = (a – b + c) (a + b – c) = [a – (b – c)] [a + (b – c)]  3bc = a2 – (b – c)2  3bc = a2 – b2 – c2 + 2bc  b2 + c2 – a2 = – bc 

b2  c 2  a2 1  2bc 2

 cos A  

1 ] 2

A = 120° 14. ABCD is a quadrilateral in which a circle is inscribed. STATEMENT-1 : The length of the sides of the quadrilateral can be A.P. and STATEMENT-2 : The length of tangents from an external point to a circle are equal. Sol. Answer (1) If sum of opposite sides of a quadrilateral is equal, then and only then a circle can be inscribed in a quadrilateral Statement-1 : Length of sides of quadrilateral can be in A.P. if common difference is zero. Statement-2 : True. 15. STATEMENT-1 : In ABC, if a < b sinA, then the triangle is possible. and STATEMENT-2 : In ABC

a b = sin A sin B


162



Sol. Answer (4) In ABC a b  sin A sin A sin A 

a b

sin A 

sin A sin B

sin B  1 16. Let ABCD be a cyclic quadrilateral then STATEMENT-1 : sinA + sinB + sinC + sinD = 0 and STATEMENT-2 : cosA + cosB + cosC + cosD = 0 Sol. Answer (4) In cyclic quadrilateral ABCD A + C = 180 and B + D = 180 A = 180 – C and B = 180 – D Then, sin A = sin C Or cos A = – cosC

and sin B = sin D and cos B = – cos D

Hence cos A + cos B + cos C + cos D = 0 17. STATEMENT-1 : In a triangle ABC if tanA : tanB : tanC = 1 : 2 : 3, then A = 45°. and STATEMENT-2 : If p : q : r = 1 : 2 : 3, then p = 1 Sol. Answer (3) Given, tanA : tanB : tanC = 1 : 2 : 3 tanA = k tanB = 2k tanC = 3k In ABC tanA + tanB + tanC = tanA tanB tanC 6K = 6K3 K  0  K = ± 1 K –1  K=1 tanA = 1  A = 45° Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



163

18. STATEMENT-1 : In an acute angled triangle minimum vlaue of tan + tan + tan is 3 3 . and STATEMENT-2 : If a, b, c are three positive real numbers then

a+b+c 3  abc also in a ABC, 3

tanA + tanB + tanC = tanA·tanB·tanC.

Sol. Answer (1)

tan   tan  tan   (tan  · tan · tan  )1/3 Equality will hold when tan   tan  tan   5

SECTION - E Matrix-Match Type Questions 1.

Match the following Column-I

Column-II

(A) The minimum value of sec2x + cosec2x – 4 is

(p) 3

(B) The maximum value of || sinx | –4 | – 3 is

(q) 4

(C) 10 log10 3  log10 tan1  log10 tan 2  .....  log10 tan 89

(r) 1

equals (D) If cosx + cos2x = 1, then the value of 4sin2x(2 – cos2x)

(s) 0

Sol. Answer A(s), B(r), C(p), D(q) (A) f(x) = sec2x + cosec2x – 4

 

sin2 x  cos2 x sin2 x cos2 x 4 sin2 2x

4

4

fmin = 4 – 4 = 0 (B) f(x) = ||sinx| – 4| – 3 fmax = |0 – 4| – 3 =1 (C) 10log10 3  log10 tan1  tan10 tan2   log10 tan89 = 3 + log10{(tan1° tan89° ) (tan2° tan88° ) …… (tan44° tan46°) tan45°} =3 (D) cosx + cos2x = 1 cosx = sin2x  4sin2x. (2 – cos2x) = 4sin2x  (1 + sin2x) ( sin2x + cos2x = 1) = 4(sin2x + sin4x) = 4(sin2x + cos2x) =4 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

164 2.



Match the expression on column I with the expression on column II that is equal to it. Column-I

Column-II

1  cos  1  cos 

(A)

(p) tan4 + sec4

2 (B) 1  tan  1  cot 2 

(C) 1 

(q) cosec + cot

2 tan 2 

(r)

cos 2 

(D) (1 – sin – cos)2

⎛ 1  tan  ⎞ ⎜⎜ ⎟⎟ ⎝ 1  cot  ⎠

2

(s) 2(1 – sin)(1 – cos)

Sol. Answer A(q), B(r), C(p), D(s) 1  cos  1  cos 

(A)

(1  cos )2





1  cos2  1  cos  sin 

= |cosec + cot|

(B)

1  tan2  1  cot 2  

sec 2  cosec 2

= tan2 2

⎛ 1  tan  ⎞ ⎧ 1  tan  ⎫ ⎨  tan ⎬ Also, ⎜ ⎝ 1  cot  ⎟⎠ ⎩ tan   1 ⎭

2

= tan2 (C) 1 

2 tan2  cot 2 

= (sec2 – tan2)2 + 2tan2 sec2 = sec4 + tan4 (D) (1 – sin – cos)2 = 2 – 2sin – 2cos + 2sin cos = 2(1 – sin). (1 – cos) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


3.


165

Match the expression given in column I with its extreme values given in column II. Column-I (A) If

sin2 x

+

Column-II cosec2 x

= 2, then 2sin x + 3cosec x is

(B) If sinx1 + sinx2 + sinx3 + sinx4 + sinx5 = ± 5, then

(p) 5 (q) –5

cos x1 + cos x2 + cos x3 + cos x4 + cos x5 is equal to (C) If cos2 x1 + cos2 x2 + cos2 x3 + cos2 x4 + cos2 x5 = 5, then

(r) 0

sin x1 + 2sin x2 + 3sin x3 + 4sin x4 + 5sin x5 is less than or equal to (D) If cos3 x1 + cos3 x2 + cos3 x3 + cos3 x4 + cos3 x5 + 5 = 0,

(s) 15

then sin3 x1 + 23 sin3 x2 + 33 sin3 x3 + 43 sin3 x4 + 53 sin3 x5 is equal to (t) 225 Sol. Answer A(p, q), B(r), C(q, r), D(r) (A) sin2x + cosec2x = 2 sinx = ± 1 = cosecx = ± 1  2 sinx + 3 cosecx = ± 5 (B) sinx1 + sinx2 + …+ sinx5 = ± 5  Each of x1, x2, ……, x5 is = ±

 2

 cosx1 = cosx2 = ……= cosx5 = 0  cosx1 + cosx2 + ……+ cosx5 = 0 (C) cos2x1 + cos2x2 +……+ cos2x5 = 5 cos2x1 = cos2x2 =……= cos2x5 = 1  cosx1 = cosx2 =……= cosx5 = ± 1 Then sinx1 + 2sinx2 + ……+ 5 sinx5 = 0 cos3x1 + cos3x2 + ……+ cos3x5 = – 5 cosx1 = cosx2 = ……= cosx5 = – 1  sinx1 = sinx2 =……= sinx5 = 0  sinx1 + 2sinx2 +……+ 5sinx5 = 0 4.

Match the following : Column-I (A)

Column-II

3 cosec 20  sec 20

 3 5 7 ⎞ ⎛ + sin4 + sin4 (B) 2 ⎜ sin4 + sin4 8 8 8 8 ⎟⎠ ⎝ (C) cos

2 4 6 cos cos 7 7 7

(D) sec10° 

1 3

cosec10°

(p) 

4 3

(q) 3

(r) 4

(s)

1 8


166



Sol. Answer A(r), B(q), C(s) D(p)

3 cosec 20  sec 20

(A)

⎛ 3 ⎞ 1 ⎜ 2 cos 20  2 sin 20 ⎟  4⎜ ⎟ 2 sin 20 cos 20 ⎟ ⎜ ⎝ ⎠

 4.

sin 40 sin 40

=4

 3 5 7 ⎞ (B) 2 ⎛⎜ sin4  sin4  sin4  sin4 ⎟ ⎝ 8 8 8 8⎠  3 ⎞ ⎛  4 ⎜ sin4  sin4 ⎟ ⎝ 8 8⎠ ⎛⎛  ⎜ ⎜ 2 sin2 ⎝⎝

2 2 2 ⎞ ⎞ ⎞ ⎛ ⎟⎠  ⎜⎝ 2 sin ⎟⎠ ⎟ 8 8 ⎠

2

⎞ 3 ⎞ ⎛ ⎛  ⎜ 1  cos ⎟  ⎜ 1  cos ⎟ ⎝ ⎝ 4⎠ 4⎠ 2

1 ⎞ 1 ⎞ ⎛ ⎛  ⎜1  ⎟  ⎜⎝ 1  ⎟ ⎝ 2⎠ 2⎠

 1

2

2

1 1  1 2 2

=3 (C) cos

2 4 6 .cos .cos 7 7 7

 2 4   cos .cos .cos 7 7 7 



1

 2 sin 7 3

8 7

1 8

(D) sec10 



.sin

1 3



1 3

cosec10

3 sec10  cosec10











167

⎛ 3 sec10  cos10 ⎞ ⎜ sin10 cos10 ⎟⎠ 3 ⎝

1

⎛ 3 ⎞ 1 ⎜ 2 sec10  2 cos10 ⎟  4⎜ ⎟ sin 20 3 ⎜ ⎟ ⎝ ⎠

1

4



3

sin(10  30) sin20

4

 5.


3

Match the value of the expression given in column-I with value greater than or equal to the number given in column-II. Column-I

Column-II

(A) 2sinx + 2cosx

(p) 2

(B) 2sin

2

x

 2cos

2

x

(q) 23/2

(C) 2sin

6

x

 2cos

6

x

(r)

(D) 2sin

4

x

 2cos

4

x

(s) 25/4

211

2

(t) 29/8 Sol. Answer A(r), B(q), C(t), D(s) (A)

2sin x  2cos x 2 2

2sin x  2cos x  2

⇒ 2 2

2

2sin

x

1 2

 2cos 2

⇒ 2sin 1

 22 2

sin x  cos x  2 2

2 2 2

1

(B)

sin x  cos x 2

2

x

2

x

 2sin x cos x

 2cos

2

x

 2sin x cos x 1

sin 2 x 1

1  1 2

3

 22


168


(C) 2sin



x

 2cos

6

1 (sin 2 x )3 1 8 2

(D) 2sin

2 6.

6

4

x

 2cos

1 (sin 2 x )2  1 4

4



x

 x


3

x cos3 x 1

2

x cos2 x 1

 2sin 9 28

 2sin

5 24

Match the following : Column-I (A) If

Column-II

sin  1 cos  3 = and = ;  and  are acute angles, sin  2 cos  2

(p) 12

then tan is equal to

sin  1 cos  3 = and = ;  and  are acute angles, sin  2 cos  2

(q)

1 5 3 3

(C) The numerical value of 4 sin50°  3 tan50° is equal to

(r)

5 3

(D) The minimum value of 9tan2 + 4cot2 is greater than or equal to

(s) 1

(B) If

then tan is equal to

(t) 2 Sol. Answer A(q), B(r), C(s), D(p) (A)

sin  1 cos  3  ,  sin  2 cos  2 sin = 2sin

cos  

2 cos  3

1  4 sin  

4 cos2  9

⇒ sec 2   4 sin2  

4 9

⇒ 1  sin2   4 sin2   ⇒ 3 sin2   1  ⇒ sin2  

4 9

4 5  9 9

5 27



⇒ tan  

(B)

5 3 3




169

1 5 3 3

sin  1 cos  3  ,  sin  2 cos  2 ⇒ sin  

1 3 sin , cos   cos  2 2

1 2 9 sin ,  cos2  4 4

1

⇒ sec 2  

1 2 9 sin   4 4

⇒ 1  sin2  

⇒

1 2 9 sin   4 4

3 2 5 sin   4 4

⇒ sin2  

5 3

⇒ sin  

5 3

(C) 4 sin50  3.sin50  4 sin50  3

sin50 cos50



4 sin50 cos50  3 sin50 cos50



2sin100  3 sin50 cos50



2sin100  2sin50 cos50



2sin100  sin80  sin 20 cos50



sin100  sin 20 2cos 60.sin 40  cos50 cos50

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170 7.




Column-II

(Trigonometric equation)

(Family of solutions)

⎛ ⎞ (A) sin 9  cos⎜   ⎟ 2 ⎝ ⎠

(p) (2n  1)

⎛ ⎞ (B) sin 5  sin⎜  2 ⎟ 2 ⎝ ⎠

(q)

n   ( 1)n , n  Z 2 4

(C) cos11 = cos3

(r)

n , nZ 7

(D) 3tan(– 15°) = tan(+ 15°)

(s) ( 4n  1)

 , nZ 10

 , nZ 14

Sol. Answer A(p), B(s), C(r), D(q)  (A) sin9 = cos ⎛⎜ – ⎞⎟ ⎝2 ⎠

 sin9 – sin = 0  cos5 = cos

 5 = n +

 2cos5 sin4 = 0

 or sin 4  0 2

n   , n Z   = (2n + 1) or   4 2 10

⎛ ⎞ (B) sin5 – sin ⎜⎝  2⎟⎠ = 0 2

⎛ 7   ⎞ ⎛ 3 –  ⎞ 2 ⎟ sin ⎜ 2⎟ ⎜  2cos ⎜ 2 2 ⎟ ⎜ ⎟⎠ = 0 ⎝ ⎠ ⎝



7  

2  n   2 2

  = (4n + 1)

 14

(C) cos11 – cos3 = –2 sin7 sin 4 = 0 =

n n or   7 4

⎡ tan  – tan15 ⎤ tan   tan15 (D) 3 ⎢ ⎥ = 1– tan  tan15 ⎣ 1  tan  tan15 ⎦ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



171

2[tan + tan2 15° tan] = 4[tan2 tan15° + tan15°]  tan sec215° = 2tan15° . sec2  sin cos = 2 sin15 cos15 

sin2 1  2 2

2 = n + (–1)n    8.

 2

n   ( 1)n , n  Z 2 4


Column-II

⎡ 3 ⎤ (A) The number of solutions of sin3x + cos2x = 0 in ⎢0, ⎥ is ⎣ 2⎦

(p) 1

(B) The number of solutions of the equation

(q) 4

⎡  ⎤ (1 – tanx)(1 + sin2x) = 1 + tanx in ⎢ , ⎥ is ⎣ 2 2⎦

(C) The number of solutions of the equation sin(x) = ex + e–x is

(r) 2

(D) The number of solutions of the equation

(s) 0

⎡  ⎤ sin3x – 3sinxcos2x + 2cos3x = 0 in ⎢ , ⎥ is ⎣ 4 4⎦

Sol. Answer A(q), B(r), C(s), D(p)

2p 3

y sin3x (A)

0 –1

p 6 p 4

p p 3 2

7p p 3p 5p 2 6

6

p 3

3p 2

x

–cos2x It is clear from the graph that 4 points satisfy the given equation (B) (cosx – sinx) (1 + sin2x) = cosx + sinx  –sinx + cosx – sinx sin2x + cosx sin2x = sinx + cosx  –1 + sin2x + 2cos2x = 1 or sinx = 0  – sinx cosx = 1 – cos2x  sinx [cosx + sinx] = 0  x = 0 or 

 4


172



(C) –1 sin(x)  1 But ex + e– x  2 Hence no solution exists. (D) sin3x – 3sinx cos2x + 2cos3x = 0  sin2x (sinx – cosx) + sinx cosx (sinx – cosx) –2 cos2x (sinx – cosx) = 0.  (sinx – cosx) (sinx – cosx) (sinx + 2cosx) = 0  (sinx – cosx)2 (sinx + 2cosx) = 0

9.

 x

 4

 x

 ⎡  ⎤   , 4 ⎢⎣ 4 4 ⎥⎦

or tan x  2  x  

 4


Column-II

(A) The range of k such that the equation sin8x + cos8x = k

⎡1 ⎤ (p) k  ⎢ , 1⎥ ⎣2 ⎦

admits of a solution is given by (B) The range of k such that the equation ksinx – 4cosx = k + 2

⎡1 ⎤ (q) k  ⎢ , 1⎥ ⎣4 ⎦

admits of a solution is given by (C) The range of k such that the equation sin4x + cos4x = k

(r) k  (–, 3]

admits of a solution is given by (D) The range of k such that the equation sin6x + cos6x = k

⎡1 ⎤ (s) k  ⎢ , 1⎥ ⎣8 ⎦

admits of a solution is given by Sol. Answer A(s), B(r), C(p), D(q) (A)

sin8 x  cos8 x  sin8 x cos8 x 2



⇒





1 2



sin4 2 x k 1   2 16 16

k

1 8

⎡1 ⎤  k  ⎢ , 1⎥ ⎣8 ⎦ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



173

(B) In order that the given equation admits of a solution we must have (k + 2)2  k2 + 16  k2 + 4k + 4  k2 + 16  k  3  k  (–3] (C) Do as ‘a’ part (A.M.  G.M) (D) Do as ‘a’ part (A.M.  G.M) 10. Match the following Column-I (A) cosx sin3x < sinx cos3x, 0  x  2 is satisfied

Column-II

⎡  5 ⎤ (p) ⎢ , ⎣ 6 6 ⎥⎦

for x lying in (B) The value of x for which 4 sin2x – 8 sinx + 3  0,

3 ⎤ ⎡   ⎤ ⎡ (q) ⎢,  ⎥  ⎢ , ⎥ ⎣ 4 ⎦ ⎣ 4 4⎦

where x[0, 2] lies in

⎡ 3 ⎤ ⎢⎣ 2 , 2 ⎥⎦  0

(C) |tan x|  1 and x [–, ] is

(r)

(D) cos x – sin x  1 and 0  x  2 is

⎛ ⎞ (s) ⎜ 0, ⎟ ⎝ 4⎠ (t)

⎡ 3 ⎤ ⎢⎣ 4 ,  ⎥⎦

Sol. Answer A(s), B(p), C(q, t), D(r) (A) cos x sin3 x – sin x cos3 x  0 sin x cos x(sin2 x – cos2 x)  0 sin x cos x(cos2 x – sin2 x)  0 sin x cos x(cos 2x)  0 sin x cos x(cos 2x)  0 sin 2x cos 2x  0 sin 4x  0 4x  [0, ]

⎡ ⎤ x  ⎢0, ⎥ ⎣ 4⎦ (B) 4 sin2 x – 6 sinx – 2 sinx + 3  0 2 sinx (2 sinx – 3) – 1(2 sinx – 3)  0 (2 sinx – 3) (2 sinx – 1)  0 But 2 sinx – 3  0 2 sinx – 1  0

sinx 

1 2

⎡  5 ⎤ x ⎢ , ⎣ 6 6 ⎥⎦ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

174



(C) | tan x |  –1  –1  tan x 2 1

3 ⎤ ⎡  ⎤ ⎡ x  ⎢  , ⎥  ⎢ ,  ⎣ 4 4⎦ ⎣ 4 ⎥⎦ (D) cos x – sin x  1

 ⎞ ⎛ ⇒ 2 ⎜ cos x sin  sin x cos ⎟  1 ⎝ 4 4⎠ 1 ⎛ ⎞ ⎡ 3 ⎤ sin ⎜  x ⎟  ⇒ x  ⎢ , 2 ⎥   0  ⎝4 ⎠ ⎣2 ⎦ 2

11. Based on the relation between the variables in column I match the type of the triangle in column II. Column-I

Column-II

(A) r1 = r2 + r3 + r

(p) Isosceles or right angled

(B) (a2 + b2) sin(A – B) = (a2 – b2) sin(A + B)

(q) Obtuse but not necessarily isosceles

(C) tanA tanB < 1

(r) Right angled but not necessary isosceles

(D) R = 2r

(s) Equilateral

Sol. Answer A(r), B(p), C(q), D(p) (A) r1 = r2 + r3 + r  r 1 – r = r 2 + r3

       s  a s s  b (s  c ) a 2s  (s  c )  s(s  a ) (s  b )(s  c )  s(s – a) = (s – b) (s – c) 

(s  b )(s  c ) 1 s (s  a )

tan2

A 1 2

A   ⇒ A  90 2 4 Then triangle is right angled (B) (a2 + b2) sin(A – B) = (a2 – b2)sin(A + B) 

sin( A  B ) a 2  b 2  sin( A  B ) a 2  b 2




175

By using componendo and dividendo 

sin( A  B )  sin( A  B ) 2a 2  sin( A  B )  sin( A  B ) 2b 2

⇒

2sin A cos B a2  2cos A sin B  b 2

⇒

sin A cos B sin2 A   cos A sin B sin2 B

⇒

cos B sin A  cos A sin B

sin2A = sin2B  A  B or 2A =  – 2B AB 

 2

Then ABC is isosceles or right angled triangle (C) tanA. tanB < 1 Let B is acute angle then  tanA < cot B  tanA < cot B (/2 – B)

AB 

 2

 C   2 (D) R = 2r ∵ cos A  cos B  cos C  1  r  3 R 2 Then it is possible A = B = C 12. Match the following Column-I (A) In triangle ABC, the angle A = 60° and

Column-II (p)

12 5

(q)

7 2

the sides b = 3 cm, c = 5 cm, then the length of median through A is (in cm) (B) In triangle ABC, the angle A = 120° and the sides b = 4 cm, c = 6 cm, then the length of angle bisector through A is (in cm) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

176



(C) In triangle ABC, the medians through A and B

are perpendicular, then the value of

(r) 8

a2 + b2 c2

(D) If the lengths of median of a triangle are 3, 4, 5 cm,

(s) 5

then the area of the triangle is (in cm2) Sol. Answer A(q), B(p), C(s), D(r) (A) A = 60°, b = 3 and c = 5

 cos A 

b2  c 2  a2 1  2bc 2

b2 + c2 – a2 = bc 25 + 9 – a2 = 15 a2 = 19 Length of median AD =

=

1 16  50  19 2

=

7 2

1 2b 2  2c 2  a 2 2

(B) A = 120°, b = 4 and c = 6 cm A ⎛ 2bc ⎞ AD  ⎜ cos ⎝ b  c ⎟⎠ 2

48 1 12   10 2 5 A

(C) B

C

D

(D) Let AD, BE and CF are the median of triangle ABC A E

F

Let B

D

C

 AD = 3 BE = 4 CF = 5 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



177

13. Match the items of Column I with Column II. Column-I

Column-II

(A) If in ABC, sin2A + sin2B = sin2(A + B),

(p) Right angled triangle

then the triangle must be (B) In a ABC,

bc = b 2 + c 2  2bc cos A , 2cos A

(q) Equilateral

then the ABC must be (C) In ABC, tan

A B C + tan + tan = 3 , 2 2 2

(r)

Isosceles

then  must be (D) In a ABC, the sides and the altitudes

(s) Obtuse angled

are in A.P. then  must be (t)

Acute angled

Sol. Answer A(p), B(r), C(q, r, t), D(q, r, t) (A) In ABC sin2A + sin2B = sin2(A + B)  sin2A + sin2B = sin2C  By using Sine Rule a2 + b2 = c2 Triangle must be right angled triangle (B)

bc  b 2  c 2  2bc cos A 2cos A By using Cosine Rule

⎛ b2  c 2  a2 ⎞ 2 2 2  b  c  bc ⎜ ⎟ 2bc 2(b 2  c 2  a 2 ) ⎝ ⎠ bc(2bc )

⇒

b2c 2 b2  c 2  a2

 a2

 b2c2 = a2b2 + a2c2 – a4  (b2c2 – a2b2) + a4 – a2c2 = 0  b2(c2 – a2) + a2(a2 – c2) = 0  (a2 – b2) (a2 – c2) = 0 a2 – b2 = 0 or a2 – c2 = 0 a = b or a = c Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

178


(C) tan


A B C  tan  tan  3 2 2 2

…(i)

A B B C C A  tan  tan  tan  tan .tan  1 2 2 2 2 2 2

∵ tan

From equation (i) By squaring both side A B B C C A⎞ ⎛ 2 A 2 B 2C  tan 2  tan 2  tan 2  2 ⎜⎝ tan 2 tan 2  tan 2  tan 2  tan 2  tan 2 ⎟⎠  3 2

 tan

A B C  tan2  tan2  1 2 2 2 2

2

A B⎞ B C⎞ C A⎞ ⎛ ⎛ ⎛ Now, ⎜ tan  tan ⎟  ⎜ tan  tan ⎟  ⎜ tan  tan ⎟ ⎝ ⎝ ⎝ 2 2⎠ 2 2⎠ 2 2⎠

2

⎡ A B C ⎛ A B B C C A⎞ ⎤  2 ⎢ tan2  tan2  tan2  ⎜ tan  tan  tan  tan  tan  tan ⎟ ⎥ 2 2 2 ⎝ 2 2 2 2 2 2⎠⎦ ⎣

= 2[1 – 1] = 0 It is possible if

tan

A B C  tan  tan 2 2 2

A = B = C (D) Let a, b, c are the sides with usual rotations given a, b, c are in A.P. 2b = a + c

A

…(i)

E

F

Let P1, P2, P3 are the length of altitudes AP, BE, CE respectively

1 1 1 Now,   aP1  bP2  cP3 2 2 2 Or P1 

B

D

C

2 2 2 , P2  , P3  a b c

And given P1, P2, P3 are in A.P. Then 2P2 = P1 + P3 ⎛ 2 ⎞ 2 2   2⎜ ⎝ b ⎟⎠ a b

2 1 1   b a b

…(ii)

From (i) and (ii) a, b, c are in A.P. as well as in H.P. then it is possible if a = b = c Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



179

SECTION - F Integer Answer Type Questions 1.

If 5cos  = 3, then,

cosec   cot  is equal to cosec   cot 

Sol. Answer (4) Given, 5cos = 3

cosec  cot  cosec  cot  = (cosec + cot)2 =

(1  cos )2 sin2 

3⎞ ⎛ ⎜⎝ 1  5 ⎟⎠ = 16 25

2.

2

4

⎛ ⎞ Least integral value of 3(tan2   cot 2 )  8(tan   cot )  10,  ⎜ 0, ⎟ can be ⎝ 2⎠

Sol. Answer (0) ⎛ ⎞ f() = 3(tan2 + cot2) – 8(tan + cot) + 10,   ⎜⎝ 0, 2 ⎟⎠

= 3(tan + cot)2 – 6 – 8 (tan + cot) + 10 2

1 1 ⎧ ⎫ ⎧ ⎫ ⎬  8⎨ ⎬4 = 3⎨ ⎩ sin  cos  ⎭ ⎩ sin  cos  ⎭

=

12

16 4 sin 2 sin 2 2

= 4(3cosec22 – 4cosec2 + 1) 4 4 ⎤ 16 ⎡ 2 4 = 12 ⎢cosec 2  cosec2  ⎥  3 9⎦ 3 ⎣ 2

2⎞ 4 ⎛ = 12 ⎜ cosec2  ⎟  ⎝ 3⎠ 3

fmin = 0 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

180


3.

x 2  2 x  2cos2   sin2   0 , then maximum number of ordered pair (x, ) such that x  R,   [0, 2].


Sol. Answer (2) x2 – 2x + 2cos2 + sin2 = 0, x, R x2 – 2x + cos2 + 1 = 0 D = 4 – 4(cos2 + 1) = – 4 cos2  x is real 0  cos = 0 In [0, 2],  

 3 , 2 2

 x2 – 2x + 1 = 0 x = 1 

4.

⎛  ⎞ ⎛ 3 ⎞ ( x, 0)  ⎜ 1, ⎟ , ⎜ 1, ⎟ ⎝ 2⎠ ⎝ 2 ⎠

The smallest value of

8 2 p for which sin(p cos) = cos(p sin) has a solution in [0, 2] is 

Sol. Answer (4) ⎛ ⎞ sin( p cos )  sin ⎜  p sin ⎟ ⎝2 ⎠

⇒ p cos  

  p sin  2

⇒ p(sin   cos ) 

⇒ sin   cos  

 2

 2p

⎞  ⎛ ⇒ sin ⎜   ⎟  ⎝ 4 ⎠ 2 2p

 2 2p

⇒ p



1

 2 2

8 2   4  2 2

 Smallest value of p is 4. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


5.

If sec( – ), sec and sec( + ) are in A.P. then cos2  sec 2


181

 +3 , then is equal to 2

Sol. Answer (5) sec( – ), sec, sec( + )  A.P.  2sec = sec( – ) + sec( + ) 

2 1 1   cos  cos(  ) cos(  )



2 cos(  )  cos(  )  cos  cos(  )cos(  )

2 2 sin  cos   cos   cos2   sin2   cos2 – sin2 = cos2sin2  cos2(1 – cos) = sin2 ⇒

6.

sin2  

sin2  1  cos 

  cos2 2 2 2  2sin 2

4 sin2

⇒

cos2  

⇒

cos2 .sec 2

 2 2

⇒

cos2 .sec 2

 3  23  5 2

⎛ ⎞ The sum of maximum and minimum values of the expression 5cos x + 3 sin ⎜  x ⎟ + 4 is ⎝6 ⎠

Sol. Answer (8) ⎛ ⎞ 5cos x  3 sin ⎜  x ⎟  4 ⎝6 ⎠ ⎛1 ⎞ 3  5 cos x  3 ⎜ cos x  sin x ⎟  4 2 ⎝2 ⎠



13 3 3 cos x  sin x  4 2 2

Maximum value 

169 27  4 4 4


182





196 4 4

= 7 + 4 = 11 Minimum value = – 7 + 4 = – 3

Sum of maximum and minimum value = 11 – 3 =8 7.

Let PQ and RS be two parallel chords of a given circle of radius 6 cm lying on the same side of the centre. If the chords subtends angles of 72° and 144° at the centre and the distance between the chords is d, then d2 is equal to

Sol. Answer (9) OM = 6 sin18°, ON = 6sin54°

O

d = 6 sin54° – 6 sin18° = 6(cos36° – sin18°)

72°

R

⎛ 5 1 5  1⎞ ⎛ 1⎞   6⎜ ⎟  3 = 6⎜ ⎟ ⎝ 2⎠ 4 ⎠ ⎝ 4

54°

P

S

M Q

 d2 = 9

8.

If 3tan·tan = 1 then

cos(  ) is equal to cos(  )

Sol. Answer (2)

tan .tan  

9.

1 3

⇒

sin .sin  1  cos .cos  3

⇒

sin  sin   cos  cos  1  3  sin  sin   cos .cos  1  3

⇒

cos(  ) 4  2 cos(  ) 2

The number of solution of 10sinx = | x | is __________.

Sol. Answer (6)

–7 –3 –2 – 2



Trigonometric Functions 2

10. The number of solutions of (81)sin x  (81)cos

2

x

183

 30 for x  [0, 2] is equal to __________.

Sol. Answer (8) 2

x

 81cos

2

x

y

81sin 81sin

y

2

x

 30

81  30 y

 y2 – 30y + 81 = 0 y = 3, y = 27 3 4 sin

2

x

2

 31, 3 4 sin

2  sin x 

x

 33

1 3 or sin2 x  4 4

∵ x [0, 2] ∵ x

 5 7 11 , , , 6 6 6 6

or

x

 2 4  5  , , , 3 3 3 3

Number of solution = 8 11. If a, b  [0, 2] and the equation x2 + 4x + 3sin(ax + b) = 2x has atleast one solution, and the least (+ve) value of a + b is k then,

2k is equal to. 

Sol. Answer (3) x2 + 4x + 3sin (ax + b) = 2x at x = 1  5 + 3 sin (a + b) = 2  sin(a + b) = –1 Minimum positive value of (a + b) is

3 k 2

 2k  2  3  3   2 12. If sin2x – 2sinx = 1 has exactly four solutions in [0, n], then minimum value of n is equal to. Sol. Answer (4) sin2x – 2sinx = 1 has exactly four solutions in [0, n] sin2x – 2sinx – 1 = 0

sin x 

2 4 4  1 2 2

sin x  1  2, sin x  (1  2) Minimum value of n = 4 for which this equation has exactly four solutions in (0, n) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

184



13. In right angled ABC, if AB = AC, then value of Sol. Answer (2) In right angled ABC

R (value [x] denote the greatest integer of x) r

C

AB = AC A = 90°, B = 45°, C = 45° And r  4R sin

A B C  sin  sin 2 2 2

r  4R sin 45 sin

=

4R 2

sin2

A

B

45 45  sin 2 2

45 2

=

4R ⎡ 1  cos 45 ⎤ ⎢ ⎥ 2 2⎣ ⎦

=

2R ⎡ 1 ⎤ ⎢1  ⎥ 2⎣ 2⎦

r 2   1  ( 2  1) R 2

R  r

1 2 1

 2.414

⎡R ⎤  ⎢ ⎥  2 ⎣r ⎦

14. In a ABC, bc = a, a = 2, the value of 2R is equal to Sol. Answer (2) In a ABC bc = a and a = 2



abc 4R

⇒ R 

abc 4

⇒ R 

a(a ) 4  4 4



⇒ R = 1

 2R = 2 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



185

⎛ bc ⎞ 15. In a triangle ABC with usual notation bcosecB = a, then value of ⎜ ⎟ is ⎝r R⎠

Sol. Answer (2)

C

Given b cosecB = a b = a sinB 

a b  1 sin B

A

A = 90°

B

ABC is a right angled triangle then  R

Now,

a A and r  (s  a ) tan 2 2

bc  r R

bc a (s  a )  2



2(b  c ) bc = a abc a s 2

bc 2 r R

16. In a triangle ABC, 2B = A + C and b2 = ac, then

a(a  b  c ) is 3bc

Sol. Answer (1) Clearly B = 60°

1 c 2  a2  b2  2 2ac 

c2 + a2 – b2 = ac

But a, b, c are in G.P. 

b2 = ac



c2 + a2 – 2ac = 0



c = a = b

Hence

a(a  b  c ) 3a 2  2 1 3bc 3a

17. In an equilateral triangle with usual notations the value of

27r 2R is equal to r1r2 r3

Sol. Answer (4) In an equilateral triangle

r 

 s


186



 Area of equilateral triangle 

s

3 2 a 4

3a 2

r 

And

R

Now

3a 2 (2) 1  a 4(3a ) 2 3

r1 

  s a

3a 2 3 a2 3 a  r2  r3   2 ⎛ 3a ⎞ ⎛ a⎞ 4⎜ 4⎜ ⎟  a⎟ ⎝ 2 ⎠ ⎝ 2⎠

a 2a  sin A 3 27r 2R 27a2 2a 8    4 43 r1r2 r3 3 3 3a 3

18. In a triangle ABC,

c cos(A  )+a cos(C + ) is equal to b cos 

Sol. Answer (1) In ABC ∵ ccos(A – ) + a cos(C + )

= c[cosA.cos + sinA.sin] + a[cosC.cos – sinC.sin] = cos[c  cosA + a cosC] + sin[c.sinA – a.sinC] By using projection formula and sine rule = cos (b) + 0 

c cos( A – )  a cos(C  ) b cos   1 b cos  b cos 

19. Let ABC and ABC  be two non-congruent triangles with sides AB = 4, AC = AC = 2 2 and angle B = 30°. The absolute value of the difference between the areas of these triangles is [IIT-JEE 2009] Sol. Answer (4) Required area will be the shaded region in the figure

A 4 2

B

30°

22

2 45°

C

Area of ACC  

2 C 1 24  4 2




187

n ⎛  ⎞ 20. The number of values of  in the interval ⎜  , ⎟ such that   for n = 0, ± 1, ± 2 and tan = cot 5 as well 2 2 5 ⎝ ⎠

as sin2 = cos4 is

[IIT-JEE 2010]

Sol. Answer (3)

tan   cot 5 we have, ⎛ ⎞  tan   tan ⎜⎝  5⎟⎠ 2



  5  n    2

 6    

  n 2

 n  12 6

Also, ⎛ ⎞ cos 4  sin2  cos ⎜  2⎟ ⎝2 ⎠ ⎛ ⎞  4  2n   ⎜⎝  2⎟⎠ 2

Taking positive

6  2n   

 2

n   3 12

Taking negative 2  2n     n 

 2

 4

Above values of  suggests that there are only 3 common solutions.

21. The maximum value of the expression

1 2

sin   3 sin  cos   5cos2 

is

[IIT-JEE 2010]

Sol. Answer (2) We have, f ( ) 

1 2

sin   3 sin  cos   5 cos2 

Let g() = sin2 + 3sincos + 5cos2 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

188





1  cos 2 ⎛ 1  cos 2 ⎞ 3  5⎜ ⎟⎠  sin 2 ⎝ 2 2 2

 3  2cos 2 

3 sin 2 2

g ()min  3  4 

3

f ()max 

9 4

5 1  2 2

1 2 g ()min

22. Consider a triangle ABC and let a, b and c denote the lengths of the sides opposite to vertices A, B and C respectively. Suppose a = 6, b = 10 and the area of the triangle is 15 3. If ACB is obtuse and if r denotes the radius of the incircle of the triangle, then r2 is equal to [IIT-JEE 2010] Sol. Answer (3)

sin C 

3 and C is given to be obtuse. 2

 C

2 3

cos C 

a2  b2  c 2 1 36  100  c 2 ⇒  2ab 2 2  6  10

c = 14

r

2

2

⎛ 15 3  2 ⎞ ⎜ ⎟ 3 ⎝ 6  8  14 ⎠

23. The positive integer value of n > 3 satisfying the equation

1 1 1   is ⎛⎞ ⎛ 2 ⎞ ⎛ 3 ⎞ sin ⎜ ⎟ sin ⎜ ⎟ sin ⎜ ⎟ ⎝n⎠ ⎝ n ⎠ ⎝ n ⎠

[IIT-JEE 2011]

Sol. Answer (7) According to the question, 1

 sin n

 sin



1 1  2 3 sin sin n n

  2 3 3 2  sin sin  sin sin sin n n n n n n




 cos

 5 2 4   3  cos  cos   cos  cos n n n n n n

 sin

9  5   sin  sin  sin 2n 2n 2n n

 sin

9 5  sin 2n 2n



189

9  5  (2k  1), say 2n

 n

7 , k is an integer 2k  1

Let us put k = 0 Hence n = 7

SECTION - G Multiple True-False Type Questions 1.

STATEMENT-1 : sin1 > sin2 > sin3 STATEMENT-2 : sin3 < sin1 < sin2

⎛ ⎞ STATEMENT-3 : sinx1 < sinx2, x1< x2, x1, x2 ⎜0, ⎟ ⎝ 2⎠ (1) T F T

(2) F T T

(3) F T F

(4) T F F

Sol. Answer (2) Statement-I : sin1 > sin2 > sin3 (false)

From graph sin2 > sin1 > sin3 Statement-II : True as explained above Statement- III : ⎛ ⎞ True clearly from graph sinx increases is ⎜⎝ 0, 2 ⎟⎠

⎛ ⎞ sinx, sinx2 if x1 < x2 is ⎜⎝ 0, ⎟⎠ 2

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190 2.



STATEMENT-1 : sin2 6° + sin2 12° + sin2 18° +....+ sin2 84° = 7 STATEMENT-2 : tan 9° tan 27° tan 45° tan 36° tan 81° = 1

  ⎛ STATEMENT-3 : ⎜⎝ tan  cot  cosec 4 4 (1) T T T

⎞ ⎛   ⎟⎠ ⎜⎝ tan  cot  cosec 4 4 4

(2) F F T

⎞ 2  ⎟⎠  sec 4 3

(3) T T F

(4) T F F

Sol. Answer (4) Statement-I : sin26 + sin212 + sin218 + …… + sin284 = 7 L.H.S. = (sin26 + cos26) + (sin212 + cos212) + …… + (sin242 + cos242) (sin284 = cos26) & so on = 1 + 1 + …… + 1 (7 times) = 7 = R.H.S. True Statement-II : Tan9°tan27°tan45° tan36 ?tan81° = 1 L.H.S. = tan9°  tan27 ° tan36° cot9° = tan27° tan36° 1 (false) Statement-III :   ⎞ ⎛ ⎜⎝ tan 4  cot 4  cosec 4 ⎟⎠

  ⎞ ⎛ 2  ⎜⎝ tan 4  cot 4  cosec 4 ⎟⎠  sec 3

L.H.S. = (1  1  2)(1  1  2) 2 = 4  2  2  sec

3.

 false 3

STATEMENT-1 : f ( )  sin2   cos2  then f(  ) = 1 for every real value of  . STATEMENT-2 : g(  ) = sec2  – tan2  then g(  ) = 1 for every real value of  . STATEMENT-3 : f(  ) = g(  ) for every real value of  . (1) T F F

(2) T T T

(3) T F T

(4) T T F

Sol. Answer (1) Statement -I : f() = sin2 + cos2 = 1  R, true Statement -II : f() = sec2 – tan2 = 1,  R, (given statement is false) ⎫ ⎧ ∵ it is true for   R  ⎨(24  1) 2 ⎬ ⎩ ⎭

Statement-III : f() = g() for all  Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



(false) ∵ g() doesn’t contain odd multiples of

191

 in its domain 2

 Domains of f and g are different  f & g are not equal T F F 4.

STATEMENT-1 : cos  = x +

1 is possible for some real values of x. x

STATEMENT-2 : sin1° < sin1. STATEMENT-3 : The minimum value of 5cosx – 12sinx + 13 is 0. (1) T T T

(2) F T T

(3) F F T

(4) F T F

Sol. Answer (2) 5cosx – 12sinx + 13   52  122  13 =0

5.

1 1 + x2 = 2. STATEMENT-1 : tan22  is a root of the equation 2 1  x2 STATEMENT-2 : sin  = x +

1 k is possible for real value x if k  . 4 x

STATEMENT-3 : |sinnx|  n|sinx| is valid for all natural numbers n. (1) T T T

(2) T F T

(3) T F F

(4) F F F

Sol. Answer (1) x  2 1

⇒

x2  2  1 2 2

⇒

x2 3  2 2  1 1 x2  1

⇒

2

x 1

x2  1

⇒

1 x2

x 2 k



3  2 2 1 3  2 2 1



42 2 22 2



⎛ 1 2 ⎞   2⎜ ⎟  2 1 2 ⎝ 1 2 ⎠

2 2

 2

k x  k 1 4

| sin nx | n | sin x | | sin nx | 10 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

192 6.


STATEMENT-1 : sin x 


1 ⇒ tan x  p

3 STATEMENT-2 : cos x  x 

1 p2  1

, p 1

3 , x  0 has no solution in R x

STATEMENT-3 : tan2 x – sin2 x = tan2 x . sin2 x, x  R. (1) F F F

(2) F T F

(3) T T T

(4) T T F

Sol. Answer (2) Statement-1 : tan x  

3 Statement-2 : As x 

1 p2  1

3  x , hence no solution exist x

Statement-3 : false for x = (2 x  1)

7.

 , x Z 2

⎛   ⎞ a STATEMENT-1 : If ,  are different values of x satisfying a cosx + b sinx = c, (b  0), then tan⎜ ⎟ . ⎝ 2 ⎠ b ⎛ ⎞ STATEMENT-2 : sin2x + cos2x + sinx + cosx + 1 = 0 has no solution in ⎜ 0, ⎟ . ⎝ 2⎠

⎡  ⎤ STATEMENT-3 : sin   3 cos   1 when  ⎢ , ⎥ . ⎣ 6 2⎦ (1) F F F

(2) T T T

(3) T T F

(4) F T T

Sol. Answer (4) Statement-1 : asinx + bsinx = c

∵ ,  are two different values of x satisfying this equation

a cos   b sin   c ∵

a cos   b sin   c a(cos   cos )  b(sin   sin )  0

 2a sin  tan

   2 2 .sin  2b cos .sin 0 q 2 2 2

 b  2 a

(∵   ) 

(False)

Statement-2 :

sin   3 cos   1 ⎛ ⎞ 1 3 .cos ⎟ L.H.S. = 2 ⎜ sin .  2 2 ⎝ ⎠ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



193

⎞ ⎛ = 2sin ⎜⎝   ⎟⎠ 3

⎞ ⎛  2 sin ⎜⎝   ⎟⎠  1 3 ⎞ 1 ⎛ sin ⎜   ⎟  ⎝ 3⎠ 2

  5    6 3 6 

   6 2

(True)

Statement-3 :

⎛ sin2x + cos2x +sinx + cosx + 1 = 0, x ⎜⎝ 0,

⎞ ⎟ 2⎠

2sinx.cosx + 2cos2x – 1 + sinx + cosx + 1 = 0 sinx(2cosx + 1) + cosx(2cosx + 1) = 0

⎛ ∵ x  ⎜ 0, ⎝

⎞ ⎟ 2⎠

 Each term is positive.  No solution 8.

(True)

STATEMENT-1 : The number of solutions of | cosx | = sinx x  [0, 4] is 4. STATEMENT-2 : The equation sin2 x 

1 3 ( 3  1)sin x   0 has two roots in 4 2

⎡ ⎤ ⎢0, 2 ⎥ . ⎣ ⎦

STATEMENT-3 : The number of solutions of sin + sin5 = sin3,   [0, ] is 5. (1) T T T

(2) T F F

(3) T T F

(4) F T T

Sol. Answer (2) Statement-1 : |cosx| = sinx Number of solutions = 4

(True)

Statement-2 : sin2 x 

sin x 

1 2

( 3  1)sin x 

3 0 4

3 1 3  1 2 3   3 2 2 2


194


=

=


3 1  2 2 2 3 1 33 , sin x  2 2 2 2

⎞ ⎟ 2⎠

⎛  One solution in ⎜⎝ 0,

(False)

Statement-3 : sin   sin5  sin3,  [0, ] 2 sin3.cos 2  sin3

sin3  0,   0,

cos 2 

 2 , , , 3 3



1 2  5 , 6 6

 6 Solution

9.

(False)

STATEMENT-1 : In a ABC, maximum value of sinA sinB sinC is

3 3 8

STATEMENT-2 : In a ABC, sin2A + sin2B + sin2C = 2 – cos2C + cosC cos(A – B) STATEMENT-3 : In a ABC, sin2A + sin2B + sin2C  9/4 (1) T T T

(2) T F T

(3) T F F

(4) F T F

Sol. Answer (1) Statement-1 : In a ABC  sin A  sin B  sin C 

3 3 8

Statement-2 : sin2A + sin2B + sin2C = 2 – cos2C + cosC.cos(A – B) R.H.S. = 1 + (1 – cos2C) + cos[ – (A + B)]cos(A – B) = 1 + sin2C – cos(A + B).cos(A – B) = 1 + sin2C – [cos2A – sin2B] = (1 – cos2A) + sin2B + sin2C = sin2A + sin2B + sin2C Statement-3 : In a ABC sin2 A  sin2 B  sin2 C 

9 4




10. STATEMENT-1 : In a ABC, if 2a2 + 4b2 + c2 = 4ab + 2ac, then cosA =

STATEMENT-2 : In a ABC if cosA =

195

1 4

1 5 , then (a + b + c)(b + c – a)  bc 4 2

STATEMENT-3 : For any ABC, the expression (b + c – a)(c + a – b) (a + b – c) – abc is negative. (1) T T T

(2) F T T

(3) T F T

(4) T T F

Sol. Answer (1) Statement-1 : In a ABC If 2a2 + 4b2 + c2 = 4ab + 2ac  a2 + c2 – 2ac + a2 – 4ab + 4b2 = 0  (a – c)2 + (a – 2b)2 = 0 It is possible of if a – c = 0 and a – 2b = 0  c  a and b 

Let a  x, b 

 cos A 

a 2

x , and c  x 2

b2  c 2  a2 2bc

x2  x2  x2 4 = ⎛ x⎞ 2 ⎜ ⎟ (x) ⎝ 2⎠ cos A 

1 4

Statement-2 : In a ABC Given cos A 



1 b2  c 2  a2  4 2bc

bc  b2  c 2  a2 2

Now, (b + c + a) (b + c – a) = (b + c)2 – a2 = b2 + c2 – a2 + 2bc Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

196


=

bc  2bc 2

=

5bc 2


Statement-3 : In a ABC (b + c – a) (c + a – b) (a + b – c) ?abc Let x = b + c – a y=c+a–b z=a+b–c Now,

xy yz 2 x  c,  a, b 2 2 2

By using A.M.  G.M x  y  2 xy y  z  2 yz

z  x  2 zx And(x + y) (y + 2) (2 + x) 8xyz (2a) (2b) (2c) 8(b + c – a) (c + a – b) (a + b – c)  (b + c – a) (c + a – b) (a + b – c) abc

SECTION - H Aakash Challengers Questions

1.

If tan  

a and 0 <  < , then the value of b

a2  b2 |a|

sin  is ....

(Assume b  0, a  0, a, b  R) Sol. tan  

a a2  b2 , sin   ?,   (0,  ) b |a|

∵ lies in (0, )  tan will be positive or negative but sin will be positive  sin  

|a| a2  b2

2 2  a  b sin   1 |a|



2.


197

If cos2009 x1  cos2009 x2  cos2009 x3  .....  cos2009 x2009  2009 , then

sin2009 x1  sin2009 x2  sin2009 x3  .....  sin2009 x2009 is equal to Sol. Given equation can be true only if cosx1 = cosx2 =……= cosx2009 = 1  sinx1 = sinx2 =……= sinx2009 = 0  sin2009x1 + sin2009x2 +……+ sin2009x2009 = 0 3.

If a cos3   3a cos .sin2   m , a sin3   3a cos2  sin   n , then (m  n )2/3  (m  n )2/3 is equal to (1) 2a2

(2) 2a2/3

(3) a2/3

(4) 2a3

Sol. Answer (2) m + n = a{(sin3 + cos3) + 3sincos(sin + cos)} = a(sin + cos) [1 – sin cos + 3sin.cos] = a(sin + cos)3 Also m – n = a[(cos3 – sin3)] + 3 sincos –(sin – cot] = a(cos – sin) [1 + sin cos – 3sin cos] = a(cos – sin)3

(1)

 (m + n)2/3 + (m – n)2/3 = a2/3[(sin + cos)2 + (cos – sin2] = 2a2/3 4.

If  satisfies 15 sin4 + 10 cos4 = 6, then find the value of 27sin8 + 8cos8

Sol. 15 sin4 + 10cos4 = 6 15 sin4 + 10 = 6(1 + tan4 + 2tan2)  9tan4 – 12tan2 + 4 = 0 (3tan2 – 2)2 = 0 2  tan  

2  sin  

2 3

2 3 & cos2   3 5

Hence 27sin8 + 8cos8  = 27 

16 8  81  625 625

=

216 (2  3) 625

=

216 125


198 5.



Let f and g be function defind by f() = cos2 and g() = tan2. Suppose  and  satisfy 2f() – g() = 1, then the value of 2f() – g() is ........

Sol. f() = cos2, g() = sin2

2f() – g() = 1 2cos2 – sin2 = 1

2cos2 = 1 + sin2 = sin2 2cos2 = sec2

2f() – g() = 2cos2 – sin2 = sec2 – sin2 =1 a ,  being an acute angle. Denote by f(a, b) the value of the expression acosec2 – bsec2, where b  = 6. Then f(8, 15) is equal to ........

Let tan  

2

a b

b

Sol. tan  

 acosec2 – bsec2

tan  

a b

a(cos 2  b sin2)  sin2.cos 2

a2 +

6.

a

 b

⎛ ⎞ a b a2  b2 ⎜ cos 2  sin 2⎟ ⎝ a2  b2 ⎠ a2  b2 sin 2.cos 2



 2 a2  b2 

sin(  2) 2sin2.cos 2

 2 a2  b2 

sin 4 sin 4

 2 a2  b2



7.

f (8, 15)  2 82  152  2 64  225  2  17  34 ⎛ ⎝

The value of 2 ⎜ cos

⎛ ⎝

Sol. 2 ⎜ cos

 3 5 7 9 ⎞ + cos + cos +cos + cos ⎟ is ........ 11 11 11 11 11 ⎠

 3 5 7 9 ⎞  cos  cos  cos  cos ⎟ 11 11 11 11 11 ⎠




199

2 ⎞ ⎛ sin ⎜ 5  ⎝ 11  2 ⎟⎠ 2 1 ⎞ ⎛  2  cos ⎜  4.  ⎟ ⎝ 11 11 2 ⎠ ⎛ 2 1 ⎞  sin ⎜ ⎝ 11 2 ⎟⎠ ⎛ 5 ⎞ ⎛ 5 ⎞ 2sin ⎜ ⎟ cos ⎜ ⎟ sin 10 ⎝ 11 ⎠ ⎝ 11 ⎠ 11  1     sin sin 11 11 8.

The value of cot16°cot44° + cot44°cot76° – cot76°cot16° is ........

Sol. cot16° · cot44° + cot44° · cot76° – cot76° cot16°

= cot16° · cot(60° – 16°) + cot(60° – 16°) · cos(60° + 16°) – cot(60° + 16°) · cot16° 2 2 ⎛ cot 60 cot16  1⎞ cos 60  sin 16 cot 60 · cot16  1  cot16 · ⎜   · cot16 ⎟ ⎝ cot 60  cot16 ⎠ sin2 60  sin2 16 cot 60  cot16

⎛ cot16  3 ⎞ 1 4sin2 16 cot16  3 · cot16  (1 3 cot16)(cot16  3 )  cot16 ⎜  ⎟ ⎝ 1 3 cot16 ⎠ 3  4sin2 16 1 3 cot16 ⎛ (cot16  3 )(1 3 cot16)⎞ 1 4sin2 16  (cot16) ⎜ ⎟ 1 3cot 2 16 ⎝ ⎠ 3  4sin2 16 ⎛ cot16  3 cot 2 16  3  3cot16  cot16  3  3 cot 2 16  3cot16 ⎞ 1 4sin2 16  (cot16) ⎜ ⎟ 1 3cot 2 16 ⎝ ⎠ 3  4sin2 16



cot 2 16 · 2 3(1 cot16) 1 4sin2 16  1 3cot 2 16 3  4sin2 16

=3

9.

If cos(  1 ) =

1 2 3

and sin(  2 ) =

1 3 2

, where 0 <  –  1 ,  –  2 <

 , then the value of 2

1 2 3

,

sin(  2 ) 

1 3 2

1

3

Sol. cos(  1 ) 

2

⎫⎪ 108 ⎪⎧ 6 2 sin(1  2 )⎬ is ........ ⎨cos (1  2 ) + 5 ⎪⎩ 18 ⎪⎭

 cos(1 – 2) = cos((1 – ) + ( – 2))

17

= cos( – 2) – ( – 1)

3

= cos( – 2)cos( – 1) + sin( – 1) sin( – 2) 



1

3 2 2 3



11



1

2 3 3 2

11

2

17

1


200





17  11 6 6

sin(1 – 2) = sin(( – 2) – ( – 1)) = sin( – 2)cos( – 1) – cos( – 2)sin( – 1) 





1 1 17 11    3 2 2 3 3 2 2 3 1  17  11 6 6

108 ⎧⎪17  11  2 17 11 6 (1  7  11) ⎫⎪   ⎨ ⎬ 5 ⎩⎪ 216 18 6 6 ⎭⎪











1 2 28  2 17 11  1  7  11 10 10



1 28  2 17 11  2  2 17 11 10



1  30  3 10





10. Let cos( – ) + cos( – ) + cos( – ) = Sol. cos(  )  cos      cos       

3 , then the value of cos + cos + cos is ........ 2

3 2

2cos( – ) + 2cos( – ) + 2cos( – ) + 3 = 0 2cos( – ) + 2cos( – ) + 2cos( – ) + cos2 + sin2 + cos2 + sin2 + cos2 + sin2 = 0 (cos + cos + cos)2 + (sin + sin + sin)2 = 0  cos + cos + cos = sin + sin + sin = 0 11. If  and  satisfy sincos =  Sol. sin cos  

1 , then the greatest value of 2cossin is ........ 2

1 2

 2sincos = – 1  sin( + ) + sin( – ) = – 1  2cossin = sin( + ) – sin( – ) = – (– 1) =1 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



201

12. If (4cos29° – 3)(4cos227° – 3) = tanK°, then K is equal to Sol. (4cos29° – 3) (4cos227° – 3) = tank°

 (2.(1 + cos18°) – 3) (2.(1 + cos54°) – 3) = tank°  (2cos18° – 1) (2cos54° – 1) = sin12°  4cos18°cos54° – 2(cos18° + sin54°) + 1 = tank° 4cos18°cos54° – 4cos36°cos18° + 1 = tank°  4cos18°(cos54° – cos36°) + 1 = sink°  1 – 4cos18° × 2sin45°.sin9° = sink°  1  4 2 

⇒

1 2

⇒ 1

sin9 cos18  sin k  cos9 sin36  tan k  cos9

2 ⎛ 1 1 ⎞ ⎜ cos9  sin9⎟  sin x  ⎠ cos9 ⎝ 2 2

 1 – 1 + tan9° = tank°  k = 9

13. If A = tan27 – tan and B 

A  B A  B 10 sin  sin3 sin9   , then the value of is.....   AB AB 3 cos3 cos9 cos 27

Sol. Given A = tan27 – tan

= tan27 – tan9 + tan9 – tan3 + tan3 – tan  tan3  tan   tan9  tan3 

sin 2 2 sin   cos3 cos  sin3

2sin3 cos9

tan27  tan9 

2 sin9 cos 27

 A = 2B 

A 2 B



AB 22 A  B 2 1 1   3,   A  B 2 1 A  B 21 3



A  B A  B 3  7 10    AB AB 3 3



A  B A  B 10   0 AB AB 3


202



14. If A  3 cot 20  4 cos 20 and B = sin12°sin48°sin54° be such that A + B = 2, then the value of 3 + 2000 is...... Sol. Given A 

 3

3 cot 20 – 4cos 20

cos 20 – 4cos 20 sin20



3 cos 20  4 sin20 cos 20 sin20



2(sin60  cos 20)  2sin 40 sin20



2 sin 60 cos 20  2 sin 40 sin 20



sin80  sin 40  2sin 40 sin20



sin80  sin 40 sin20



2cos 60 sin 20 sin 20

 2

1 1 2

B = sin12°sin48°sin54°



1 (2 sin 48 sin12)cos36 2



1 (cos36  cos 60)cos36 2



1 ⎛ 5  1 1⎞ 5  1  ⎟ 2 ⎜⎝ 4 2⎠ 4



1 5 1 5 1   2 4 4



1 4  2 16



1 8




203

Now, A + 8B = 1 + 1 = 2 A + 8B = 2 =8 

3 + 2000 = 83 + 2000 = 512 + 2000 = 2512

1

15. If tan  

2

x( x  x  1)

, tan  

x 2

x  x 1

and tan  

1 x

3



1 x

2



1 be such that l + m + n = 0, then the x

value of l3 + m3 + n3 – 3lmn is ...... (where l, m, n  Z having no common factor except one) Sol. Given tan  

1 2

x( x  x  1)

1

tan  

1 x  x x3

2

Now, tan(  ) 

1 

2

x( x  x  1) 1

, tan  

x 2

x  x 1

x x



2

x  x 1

tan   tan  1  tan .tan 



x 2

x  x 1

x x ( x  x  1) 2

(1  x )

x 2 x  x  1  x2  x  1 1 ( x 2  x  1)







( x  1) x x2  x  1



( x 2  x  1) x ( x  1)

x2  x  1 x x 1 1 1    tan  x x2 x3


204



 +=   +  –  = 0  l = 1, m = 1, n = – 1 Now, l3 + m3 + n3 = 3lmn = 1 + 1 – 1 + 3

=4 16. A = { | 2cos2 + sin  2} ⎧  3 ⎫ B = ⎨    ⎬ . Find A  B 2 2⎭ ⎩

Sol. A = {|2cos2 + sin  2} 2 sin2   sin   0

sin (2 sin   1)  0

…(1)

Since

y

⎡  3 ⎤  ⎢ , ⎥ ⎣2 2 ⎦

…(2)

x'

Hence (1) and (2) will represent ⎡  5 ⎤ ⎡ 3 ⎤   ⎢ , ⎥  ⎢ , ⎥ ⎣2 6 ⎦ ⎣ 2 ⎦

0

⎛ 5 ⎞ ⎜ 6 ,0 ⎟ ⎝ ⎠

x

y'

17. Solve for , 4cot2 = cot2 – tan2. Sol. 4cot2 = cot2 – tan2

4(cos2   sin2 ) cos2  sin2  cos2   sin2   . sin .cos  cos  sin  sin .cos   4 cos 2.

sin 2  cos 2 2

 cos 2.(2 sin 2  1)  0  cos   0

or

   (2n  1)

 4

  (2n  1)

sin 2 

or



1 2

n   ( 1)n . 2 12

 n   ( 1)n . or 4 2 12

18. If 3tan( – 15°) – tan( + 15°) = 0. Find . Sol. 3tan( – 15) – tan( + 15) = 0

tan(  15) 3 tan(  15) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



205

Applying componendo - dividendo,

sin(  15    15) 3  1  sin(  15    15) 3  1 sin2  2 

1 2

=1 2  n  ( 1)4 .



 2

n   ( 1)4 . 2 4

1 ⎞ ⎛ 19. If cos2  ( 2  1)⎜ cos   ⎟ , then general value of  is _______. 2⎠ ⎝ 1 ⎞ ⎛ Sol. cos 2  ( 2  1) ⎜⎝ cos   ⎟ 2⎠

2cos2   1  ( 2  1)cos   1 

1 2

0

2 2 cos2   (2  2)cos   1  0

( 2 cos   1) (2cos   1)  0 cos  

1 2

  2n  

or

  cos  

1 2

  or 2n   4 3

20. Solve the equation, sin2n – sin2 [(n – 1)] = sin2. Sol. sin2n – sin2(n – 1)  = sin2  sin(2n  )  sin   sin2   sin   0 or sin(2n – 1)  = sin sin = 0  = k sin(2n – 1) = sin (2n – 1)  = k + (–1)k k = even = 2m (2n – 1) = 2m +  

m n2


206



k = odd = 2m + 1 (2n – 1 ) = (2m + 1) – 



(2m  1) 2n

⎛ 2m  1⎞    k  or ⎜⎝ ⎟ 2n ⎠

or

m n2

where k, m  Z

21. Solve the equation tan x 

cos x

Sol. tan x 

1  sin2 x

2

tan x 

cos x  2, sin x  cos x

tan x 

1 2 1  tan x

(1  tan x ) 

cos x 2. 1  sin2 x

if sin x  cos x  0

1 3 1  tan x

Let 1 + tanx = y y2 – 3y + 1 = 0  y 

3 5 2

 tan x 

tan x 

1 5 2

5 1 , 2

tan x 

 5 1 2

⎛ 5  1⎞ ⎛  5  1⎞ x  n   tan1 ⎜ or n   tan1 ⎜ ⎟ 2 ⎟⎠ ⎝ 2 ⎠ ⎝ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



207

Case II sin x + cos x < 0 tan x 

cos x 2 sin x  cos x

tan x 

1 2 1  tan x

1  tan x 

1 3 1  tan x

y 2  3y  1  0

y

(1  tan x  y )

3  13 2

tan x 

1  13 2

1  13 1  13 , 2 2

22. Let x and y, (0  x, y 

 ) satisfy 3sin 2 x + 2sin 2 y = 1 and 3sin2x = 2sin2y, then the value of 2

232 ( x  2y ) is _______. 

Sol. We have, 3 sin2x + 2 sin2y = 1 3 sin2x = cos2y and 3sin2x = 2sin2y ⎛ ⎞ tanx = cot2y = tan ⎜⎝  2y ⎟⎠ 2

x + 2y = n +

= (2n  1)

 2

Since 0 < x, y <

0 < x + 2y < 

 , 2 n Z

 2  3 x + 2y = 2 2

232 ⎛ 232 ⎞ ⎛  ⎞ ( x  2y )  ⎜  116 ⎝  ⎟⎠ ⎜⎝ 2 ⎟⎠ 


208



2 cos A  cos B  cos 3 B and

23. If angles A and B satisfy 1620sin2(A – B) is _______. Sol.

2 sin A  sin B  sin 3 B , then the value of

⎛ 3  cos 2B ⎞ 2 cos A  cos B(1  cos2 B )  cos B ⎜ ⎟⎠ ⎝ 2

⎛ 1  cos 2B ⎞ 2 ⎟⎠ 2 sinA = sinB (1 – sin B) = sin B ⎜⎝ 2

and

2 sin( A  B )   sin B cos B

Hence,

 2 2 sin( A  B )   sin2B 2

and 8 =

cos2B

⎛ 3  cos 2B ⎞ ⎛ 1  cos 2B ⎞ 2 ⎜⎝ ⎟⎠  sin B ⎜⎝ ⎟⎠ 2 2

2

After simplification we shall get cos2B =

1 2 2 and sin 2B = 3 3

Hence 1620 sin2 (A – B) = 1620 

sin2 2B 8  1620   180 8 89

24. Let the sum of all solutions of the equation 3 3 sin3 x  cos3 x  3 3 sin x cos x  1 in the interval [0, 10] be denoted by S. Then

3S is _______. 

Sol. We have, 3 3 sin3 x  cos3 x  3 3 sin x cos x  1

 



3 sin x

3

  cos x    1  3 3

3





3 sin x  cos x  1  0

3 sin x  cos x  1  0

 cosx + 



 cos x 

3 sin x  1

1 3 1  .sin x  2 2 2

⎞  ⎛  sin ⎜⎝ x  ⎟⎠  sin 6 6 ⎞  ⎛ or cos ⎜ x  ⎟  cos ⎝ 3⎠ 3

 x 

   2n  3 3



 x = 2n, 2n +


209

2 ,nZ 3

The solutions of the given equation lying in [0, 10] are 0, 2, 4, 6, 8, 10 ;

2 8 14 20 26 , , , , 3 3 3 3 3

Thus the sum of all solutions of the given equation is



160 s 3

3s 3 160    160   3

25. Solve the equation, 4sin2 x  2cos x  41  sin2 x  2sin 2

Sol. 4sin 2 x  2 cos

2

x

2

 41 sin 2 x  2 sin

x

2

x

 65 .

 65

 41 cos 2 x  sin 2 x  42  sin 2 x cos x  65  4  4cos 2 x  sin x  16.4  sin 2 x  cos 2 x  65 Let 4cos2x + sin2x = t  4t 

16  65 t

 4t2 – 65t + 16 = 0 4t2 – 64t – t + 16 = 0 4t2 – 64t – t + 16 = 0 4t (t – 16) – (t – 16) = 0 t = 16

or t 

1 4

cos2x + sin2x = 2

or cos2x + sin2x = –1

not possible Ans. x  n  

 4

or x = (2n  1)

 2

26. If the equation sin6x + cos4x = –2 have a family of nonnegative solutions xk’s, where 0  x1 < x2 < x3 _______. < xk < xk+1 ......., then the value of

1 

1000

∑| xk 1  xk | is _______. k 1

Sol. We have, sin6x + cos4x = –2 Which will be possible only when sin 6 x  1 , cos 4x = –1 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

210


 6 x  2k  

x = n 


 , 4 x  2m  , k, m  Z 2

 where n  Z . 4

So xk+1 – xk =  

1 1000 ∑ xk 1  xk  1000  k 1

27. Find all values of x, y and k for which the system of equations sinx cos2y = k4 – 2k2 + 2 cosx sin2y = k + 1 has a solution. Sol. sinxcos2y = k4 – 2k2 + 2 cosxsin2y = k + 1 sinxcos2y = (k2 – 1)2 + 1 and cosxsin2y = k + 1 But L.H.S. is less than or equal to 1.  k=±1 So, Case-I, k = 1 sinxcos2y = 1 sinx = 1 and cos2y = 1 or cosxsin2y = 2 which is not possible Case-II, When k = –1 sinx cos2y = 1 sinx = 1 and cos2y = 1 or, sinx = –1 or cos2y = –1 and cosx sin2y = 0 x  2n  

 2

and y 

28. Solve for x and y if Sol.

n 2

sin x cos y  0 and 2sin2x – cos2y – 2 = 0.

sin x cos y  0 and 2sin2 x  cos 2y  2  0 sinx = 0 or cosy = 0 when sinx = 0




211

cos2y = – 2 (which is not possible) So, cosy = 0 2sin2x – (2cos2y – 1) – 2 = 0  2sin2x = 1  sin x  

y  2n  

1 2

 2

29. The sum of value of , 0   2, such that x2 – 2xcos + 1 = 0 has real roots is ....... times . Sol. x2 – 2xcos + 1 = 0

x=

=

2cos   4 cos2   4 2

2cos   4 sin2  will be real if sin = 0 2

 A = 0, , 2  sum of values = 0 +  +2 = 3

30. If cos

 2+ 6 3 1 3 +1 ⎛ ⎞ = , then all x  ⎜ 0, ⎟ such that + = 4 2 , then find x. 12 4 sin x cos x 2 ⎝ ⎠

⎛ 3 1 3  1⎞ ⎜ ⎟ 2 2  2 2 ⎟ Sol. 2 2 ⎜ cos x ⎟ ⎜ sin x ⎜⎝ ⎟⎠





  ⎛ ⎞ ⎜⎝ sin cos x  cos sin x ⎟⎠ 12 12  2 2 sin x cos x ⎛  ⎞⎞ ⎛ ⎜ sin ⎜⎝ x  12 ⎟⎠ ⎟  4 2⎜ ⎟ sin2 x ⎟ ⎜ ⎜⎝ ⎟⎠ x

  2x 12

x

 12


212



31. The number of values of x lying in [–, ] and satisfying 2sin2 = cos2 and sin2 + 2cos2 – cos – 1 = 0 is (1) 0

(2) 2

(3) 4

(4) 6

Sol. Answer (2) 2sin2 = cos2  2sin2 = 1 – 2sin2  sin   

1 2

…(i)

sin2 + 2cos2 – cos – 1 = 0 2sin cos – cos + 2(1 – 2sin2) – 1 = 0 cos(2sin –1) + (1 – 4sin2) = 0 (2sin –1) [cos – 2sin – 1] = 0 sin  

1 or cos – 2sin = 1 …(ii) 2

From (i) & (ii) Common solution is obtained at sin  

1 2

 Number of solutions in [–, ] is 2 32. The number of solutions of the equation sinx.cosx(cosx – sinx)2. (sinx + cosx) = , where  

1 2 2

in the

interval [0, 4], is (1) 0

(2) 2

(3) 4

(4) 8

Sol. Answer (1) sin2x.(cosx – sinx). cos2x = 2

⎞ 4 ⎛  sin4 x.cos ⎜ x  ⎟  ⎝ 4⎠ 2 = 2 2 L.H.S. can never exceed one while RHS = 2 2  1 Since  

1 2 2

 No solution 33. The number of solutions of the equation sin5   (1) 0

(2) 1

1 1 ⎛   cos5  where  ⎜ 0, ⎝ sin  cos 

(3) 2

⎞ ⎟ , is 2⎠

(4) 3

Sol. Answer (2) sin5  – cos5  

1 1  cos  sin 




(sin– cos) [sin4 + sin3 cos + sin2 cos2 + sin cos3 + cos4] =

213

sin   cos  sin  cos 

 sin – cos= 0

…(i)

OR sin4 + cos4 + sincos (sin2 + cos2) + sin2 cos2– From (i) sin   cos  ⇒  

1 0 sin  cos 

…(ii)

 4

From (ii) 1 1 1 2 1 sin2 2  sin2  sin2 2  0 2 2 4 sin2

 sin32 – 2sin–22 – 4sin2 + 8 = 0  (sin2– 2) (sin22 – 4) = 0  No solution  Only solution is  

 4

34. The number of solutions of x in the interval [–, ] of the equation (1 + cot267º) (1 + tan222º) = sec2x + cos2x is (1) 1

(2) 2

(3) 3

(4) 4

Sol. Answer (3) L.H.S. = (1 + cot267°) (1 + tan222°) = (1 + tan3°) (1 + tan42°) ∵ 3 + 42 = 45° 

tan3  tan42  tan45 1 tan3 tan42

 tan3° + tan42° + tan3° tan42° = 1  (tan3° + 1) (tan42° + 1) = 2  L.H.S. = 2  Given equation is sec2x + cos2x = 2 ∵ sec2x + cos2x  2, equality colds of sec2x = cos2x  x = , 0,  (As the given interval)  3 solutions 35. The general solution of the equation tan(x + 20º) tan(x – 40º) = tan(x – 20º) tan (x + 40º) is (1) x = n, n  Z

(2) x  n  

 , n Z 4

(3) x 

n , n Z 2

(4) x  2n  

 ,nZ 4


214



Sol. Answer (3)

tan( x  20) tan( x  40)  tan( x  20) tan( x  40 A) =

sin( x  20)cos( x – 20) sin( x  40)  cos( x – 40)  sin( x – 20)cos( x  20) cos( x  40)sin( x – 40)

Applying components-dividends 

sin2 x sin(2 x )  sin40 80

 sin2x = 0  2x = n 

x

n , n Z 2

36. Let x  [0, 2]. The curve y = secx tanx + 2tanx – secx and the line y = 2 intersect in (1) No point

(2) 2 points

(3) 3 points

(4) 4 points

Sol. Answer (4) secx tanx + 2tanx – secx – 2 = 0  (secx + 2) (tanx – 1) = 0  tan x  1, sec x  2 

 5 2 4  x  , ,x  , (in [0, 2]) 4 4 3 3

The curve and the line intersect is 4 points 37. Let the number of solutions of the equation sinx = x2 – 4x + 5 be . The general solution of the equation





1  tan (  2  2  1) 

(1) n  

 only n  Z 4

2 1  tan((  2  2  1))

(2) n  

 0 is

 , n Z 4

(3) n  

  only n  Z (4) n   , n Z 3 3

Sol. Answer (4) sinx = (x – 2)2 + 1 –1 L.H.S  1 and RHS 1 L.H.S. = 1 at x 

 while R.H.S. = 1 at x = 2 2

No solution  =0 Putting  = 0 is given equation, we get 1 tan  

2 0 1 tan 

1 – tan2 + 2 = 0  tan2 = 3  General solution is   n 

 3




215

38. Let three sets be defined as A = {x|cos2x + cosx + 1 = 0} B = {x|cos2x + 3sinx = 2} C = {x|secx + tanx = cosx} The number of elements in A B C is (1) 0

(2) 4

(3) 8

(4) Infinitely many

Sol. Answer (1) (i) For set A cos2x + cosx + 1 = 0 2cos2x + cosx = 0 cosx(2cosx + 1) = 0 cosx = 0 or cosx = –

1 2

 2 x  (2n  1) or x  2n  2 3

(ii) For set B cos2x = 3sinx = 2 1 – 2sin2x + 3sinx = 2  2sin2x – 3sinx + 1 = 0 (2sinx – 1) (sinx – 1) = 0

sin x 

1 or sinx = 1 2

x  n  (–1)n 

  or x  2n  6 2

(iii) For set C secx + tanx = cosx 1 + sinx = cos2x  sin2x + sinx = 0 sinx = 0, sinx –1 as cosx 0  x = n From (i), (ii), (iii) it is clear that these exists no x satisfying all 3 conditions  No. of elements in A B C = 0 39. The number of solutions of the equation 2|x| = 1 + 2|cosx| is (1) 0

(2) 2

(3) 4

(4) Infinite


216



Sol. Answer (2) Clearly from graph these are two solution of the given equation

y

|x|

y=2

y = 1 + 2|cosx|

0

 2



x

40. If the equation 1 + sin2x = cos has a non-zero solution in , then x must be (1) An integer

(2) A rational number

(3) An irrational number (4) None of these

Sol. Answer (2) Given 1 + sin2x = cos OR cos = 1 + sin2x It is possible if sin2x = 0 and cos= 1 x = n and  = 2m where n; m I 

x

x

n 2m

n 2m

41. In a triangle ABC, if the median and altitude from A trisect angle A, then (B – C) is ( in degrees) equal to _______. Sol. From the question it is clear BD = DE and BE = EC Let BD = x So, BD = DE = x & EC = 2x  AE is the angle bisector of ADC So,

AD 1 AD  , But  sin C AC 2 DC

 sinC = sin30 C C = 30° 

AD AD 1  tan30 ⇒  ⇒ AD  3 x DC 3x 3

In ABD

tan B 

AD  x

3x x

 B = 60 So, B – C = 30° Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456



217

42. In a right angled triangle ABC, if r(inradius) = 7 cm and R(circumradius) = 32.5 cm, then the area of the triangle (in square cm) is _____ Sol. r = 7 cm R = 32.5 cm A=? Let C = x C = C5 AP = A5 [Length of tangent drawn from internal point are equal]

A

Apply Pythagoras theorem in w.r.t. ABC AB2

+

BC2

65 – x

AC2

=

7

0

Simplifying we get x = 56 or 9 Ar ABC 

65 – x

5

(72 – x)2 + (7 + x)2 = 652

7 C

1 1  BC  AB   16  63 2 2

x



7

P 7 B

= 504 sq. units 43. In a triangle ABC, perpendicular distance of BC from the point of intersection of angle bisector of B and C is _____. Sol. From figure it is clear that the perpendicular distance of BC from the point of intersection of angle bisector of B and C is = inradius =r 1

44. In a ABC if

r

2

+

1 r12

+

1 r22

+

1 r32

=

a2 + b2 + c 2 n

, then n is equal to ______.

Sol. Given that, 1 r2



1 r12



1 r22



1 r32



a2  b2  c 2 n

Taking L.H.S. 1 r





2



1 r12



1 r22



1 r32

s 2  (s  a )2  (s  b )2  (s  c )2 2 (a  b  c )2  (b  c  a )2  (a  b  c )2  (a  b  c )2 4 2


218




4(a 2  b 2  c 2 ) 4 2




a2  b2  c 2 2

So n = 2 45. The radius of the circle touching two sides AB and AC of a triangle ABC and having its centre on BC is equal to ______ .

A

Sol. Let the radius of the circle be r ar(ABC) = ar(ABD) + ar(ACD)

1 1    · r · b  · r ·c 2 2  r

r B

r D

C

2 bc

46. Find the cubic equation whose roots are the radius of three inscribed circles in term of inradius, circumradius and perimeter. Sol. Let roots be r1, r2 and r3  Equation x3 – (r1 + r2 + r3)x2 + (r1r2 + r2r3 + r1r3)x – r1r2r3 = 0 r1 + r2  r2r3 + r1r3 =

2 2 2   (s  a )(s  b ) (s  b )(s  c ) (s  c )(s  d )

⎡ s c s as b ⎤  s2 ⎢ ⎥ ⎣ s (s  a ) (s  b ) ( s  c ) ⎦ 

s  2 .s 2

 s2

1 1 1 r2 r3  r1r3 s2 1       r r Also, 12 r1 r2 r3 r1r2 r3 r1r2 r3 r  r1r2r3 = s2r and r1 + r2 + r3 = 4R + r  Equation x3 – (4R + r)x2 + s2x – s2r = 0 47. ABCD is a cyclic quadrilateral inscribed in circle of radius 21 cm, having diagonals AC and BD at right angles, the point of intersection of AC and BD being E. Then the sum of squares of E from the four vertices (in sqaure cm) is _______. A Sol. We have to find EB2 + ED2 + EA2 + EC2





1  (EB 2  EA2 )  (EB 2  EC 2 )  (EC 2  ED 2 )  (ED 2  EA2 ) 2 



1 AB 2  BC 2  CD 2  AD 2 2



D



E

90°– 

 90°–



B



C




219

Now, let R be the radius of the circumscribed circle using sine rule AD2 = 4R2 sin2 ; AB2 = 4R2 sin2 BC2 = 4R2 cos2 ; CD2 = 4R2 cos2 Hence





1 AB 2  BC 2  CD 2  AD 2 2



 

1 8R 2  4R 2  1764 2

48. In ABC, 3sinA + 4cosB = 6 and 4sinB + 3cosA = 1. Find the measure of angle C.

3 sin A  4cos B  6 Sol.

3cos A  4 sin B  1 3(sin A  cos A)  4(cos B  sin B )  5

4sinB + 3cosA = 1

9 sin2 A  16 cos2 B  24 sin A cos B  36 9 cos2 A  16 sin2 B  24 cos A sin B  1 9  16  24(sin A cos B  cos A sin B )  37

24sin(A + B) = 37 – 25 = 12 sin(A + B) =

1 2

 sin( – C) =

 sinC =

C

1 2

1 2

 5 , 6 6

49. Maximum value of tan

Sol. tan

A B C · tan · tan in a triangle is equal to 2 2 2

A B C  tan  tan 2 2 2

 Given A + B + C = 

⇒

⎛ A  B  C⎞  ⎜⎝ ⎟⎠  2 2

⇒

tan

A B B C C A tan  tan tan  tan tan  1 2 2 2 2 2 2


220





tan A tan B tan B tan C tan C tan A 3 A B C       tan2  tan2  tan2 2 2 2 2 2 2 2 2 2

⇒

1 A B C  tan2 tan2 tan2 27 2 2 2

⇒

tan

A B C 1 tan tan  2 2 2 3 3

50. If the angle A and angle B of a triangle (A > B) satisfy the equation 2tanx – (1 + tan2x) = 0, then determine the angle C where 0 <  < 1. Sol. 2tanx – (1 + tan2x) = 0 A, B satisfy this equation 2 sin x   cos x cos2 x

 sin2x =   sin2A =  = sin2B  2A =  – 2B  A  B 

 C

(∵ A  B )

 2

 2


Cls Jeead-15-16 Xi Mat Target-1 Set-2 Chapter-3

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