Trigonometry

JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 1 of A - 34

TRIGONOMETRIY TRIGONOME TRIY (COLLECTION # 1)

Single Correct Type

cos A co cos C + cos(A + B) co cos(B + C)

simplifies to Que. 1. If A + B + C = 180o then cos A si sin C − sin(A + B) co cos(B + C) (a) − cotC

(b) 0

(c) tan C

(d) cot C

(code-V1T2PAQ2)

Que. 2 Let 3a = 4, 4b = 5, 5c = 6, 6d = 7, 7e = 8 and 8f = 9. The value of product (abcdef), is (a) 1

(b) 2

(c)

6

(d) 3

(code-V1T2PAQ3)

(d) cos ec15 ec15o

(code-V1T2PAQ4)

Que. 3. Which of the following numbers is the largest ? (a) cos15ο

o (c) sec15

(b) tan60o

sin α − sin γ

simplifies to Que. 4. If α + γ = 2β then the expression cos γ − cos α (a) tan β

(b) − tan β

Que. 5. If A = 110o then

1 + 1 + tan 2 2A

(c) cot β

(d) − cot β

(c) cot A

(d) – cot A

(code-V1T4PAQ2)

(d) 776

(code-V1T4PAQ4)

equals

tan2A

(a) tan A

(code-V1T2PAQ6)

(b) – tan B

Que. 6. Minimum value of y = 256 sin 2 x + 324 cos ec 2 x ∀ x ∈ R is (a) 432

(b) 504

Que. 7. If A = 320 then o

(a) tan

(c) 580

−1 + 1 + tan 2 A

A

tanA

(b) − tan

2

is equal to

A

(code-V1T5PAQ1)

(c) cot

2

A

(d) − cot

2

A 2

x  x  x x    Que. 8. The product  cos  . cos  .  cos  .......  cos  is equal to 2  4  8 256    sinx

(a)

128sin

sinx

(b)

x 256

256sin

x 256

sinx

(c)

128sin

(code-V1T5PAQ3)

sinx

(d)

x

512sin

128

x 512

Que. 9. In a triangle ABC∠A = 60o ,∠B = 40o and ∠C = 80o. If P is the centre of the circumcircle of triangle ABC ABC with radius unity, then the radius of the circumcircle of triangle BPC is (code-V1T5PAQ5) (a) 1

(b)

3

(c) 2

(d)

3 2

Que. 10. In 10. In a triangle ABC if angle C is 90 o and area of triangle is 30, then the minimum possible value of the hypotenuse c is equal to (code-V1T5PAQ9) (a) 30 2

(b) 60 2

(c) 120 2

(d) 2 30

Que. 11. Which one of the following relations does not hold good ? (a) sin

π 10

sin

13π 10

=−

1 4

sin 24o + cos 6o ) = 15 + 3 (c) 4 ( si

THE “BOND”

|| Phy. Phy. by Chitranjan|| Chitranjan||

(b) sin

π 10

+ sin

13π 10

(code-V1T5PAQ13)

=−

2 o 2 o (d) sin 72 − sin 60 =

1 2 5 −1

||Chem. by Pavan Gubrele|| Gubrele||

4

||Maths by Suhaag Kariya||

JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 2 of A - 34 Que.12. With usual notations, in a triangle ABC, a cos ( B − C ) + b cos (C − A ) + c cos ( A − B ) is equal to (a)

abc R

abc

(b)

2

4R

(c)

2

4abc R

(d)

2

abc

(code-V1T5PAQ15)

2R 2

Que. 13. General solution of the equation, 2 sin 2 x + sin 2 2 x = 2, is (a) ( 2n + 1)

π

(b) nπ ±

4

π

π

(c) nπ ±

4

(code-V1T5PAQ16)

(d) nπ ±

2

π 4

∪ nπ ±

π 2

ABC, ∠A = 60o and b : c = 3 + 1 : 2 then ( ∠B − ∠C ) has the value equal to Que. 14. In a triangle ABC, (a) 15o

(b) 30o

(c) 22.5o

(d) 45O

(code-V1T5PAQ17)

Que. 15. If the two roots of the equation, x 3 − px 2 + qx − r = 0 are equal in magnitude but opposite sign then (a) pr = q

(b) qr = p

2 Que. 16. The equation, sin θ −

(a) no root

4 sin

θ −1

2

(c) pq = r

=1−

4 sin

3

(b) one root

θ −1

has

(c) two roots

(b) [1,2]

(code-V1T5PAQ18)

(code-V1T5PAQ20)

Que. 17. If a,b,c are the sides of a triangle then the expression (a) (1,2)

(d) pq + r = 0

(d) infinite roots a 2 + b2 + c2 ab + bc + ca

(c) [1,2)

lies in the interval

(d) (1,2]

(code-V1T5PAQ23)

Que. 18. If α and β are the roots of the equation a cos 2θ + b sin 2θ = c then cos 2 α + cos 2 β is equal to (a)

a 2 + ac + b 2

(b)

a 2 + b2

Que. 19. If tan θ =

b a

(b) b

Que. 20. Let f k ( x ) = 1 4

(c)

a 2 + b2

1

r 2R

(d)

2a 2 b2 + c2

(b)

1 12 A 2

1

+ sin 2 2 x

(c)

4

+ cos 2

(b) 4 −

B 2

+ cos 2

7r 2R

C 2

1 3

(d) a 2 b

1

− cos 2 x 4

(code-V1T7PAQ6)

(d)

1 12

equal

(c) 2 +

(code-V1T5PAQ24)

(code-V1T7PAQ5)

then f 4 ( x ) − f 6 ( x ) is equal to

k

2 tr iangle ABC, cos Que. 21. In 21. In a triangle

(a) 2 +

a 2 + c2

(c) ab2

sin k x + cos k x

− cos2 2 x 6

2b 2

, then the value of a cos 2θ + b sin 2θ is equal to

(a) a

(a)

a 2 − ac + b 2

(code-V1T7PAQ7)

r 4R

(d)

3  4r

 + 1  4 R 

where r and R have their usual meaning. Que. 22. 22. A cricle is inscribed in a triangle ABC touches the side AB at D such that AD = 5 and BD = 5 and BD = 3. If ∠A = 60 o then the length length of BC equals equals [Advise - Que. of properties of triangle but tru by by 2D.] (a) 9

THE “BOND”

(b)

120 13


(c) 12

(d) 13


(code-V1T7PAQ8)


JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 3 of A - 34 tr ianlge ABC, sin A,sin A.P., then Que. 23. 23. If in a trianlge A,sin B,sin ,sin C are in A.P.,

(code-V1T7PAQ10)

(a) The altitudes are A.P.

(b) The altitudes are in H.P

(c) The medians are in G.P.

(d) The medians are in A.P.

Que. 24. 24. Which of the following inequalitie(s) hold(s) true in any triangle ABC ? (a) sin

A 2

2 (c) sin

sin

A

π 7

2

sin 2

2

Que. 25. 25. If α =

B

sin

B 2

C 2

≤

sin 2

1

(b) cos

8

C 2

<

3

A 2

2 (d) cos

4

cos

A 2

B 2

cos

+ cos 2

B 2

C 2

≤

(code-V1T7PAQ13)

3 3 8

+ cos 2

C 2

≤

9 4

which of the f ollowing ollowing hold(s) good ?

(code-V1T7PAQ14)

(a) tan α. tan 2α. tan 3α = tan 3α − tan 2α − tan α (b) cos ec α = cos ec 2α + cos ec 4α (c) cos α − cos 2α + cos 3α has the value equal to 1/2 (d) 8 co cos α. cos 2α. cos 4α has the value equal to 1. Que. 26. 26. Identify which of the following are correct ? (a) ( tan x )

n ( sin x )

> ( cot x )

 n( sin x )

 

, ∀ x ∈  0,

(code-V1T7PAQ16)

π 

4

π cos ec x (b) 4n co < 5n cosec x , ∀ x ∈  0,   2 1 (c)   2

n ( cos x )

 1 <   3

n ( cos x )

 

, ∀ x ∈  0,

π 

2

(d) 2n ( tatan x) < 2n( sisin x) , ∀ x ∈  0, π   2 Que. 27. 27. Which of the following is always equal to cos2 A − sin 2 A ?

(code-V1T10PAQ4)

(a) sin2A c os ( A − B) − sin ( A + B) si s in ( A − B ) (b) cos ( A + B ) co c os ( A − B ) − cos ( A + B) si s in ( A − B ) (c) sin ( A + B ) co s in ( A − B ) − sin ( A + B) co c os ( A − B) (d) cos ( A + B ) si

Que. 28. 28. Number of values of x ∈ [0, π] satisfying cos 2 5x − cos 2 x + sin 4x 4x. sin 6x 6x = 0, is(code-V1T10PAQ5) (a) 2

(b) 3

(c) 5

(d) infinitely many

sin α + sin β + sin ( α + β)

Que. 29. 29. sin α + sin β − sin α + β (wherever difined) simplifies to ( ) (a) cot

α 2

cot

THE “BOND”

β 2

(b) cot

α 2

tan

β 2


(c) tan

α 2

cot

β 2

(code-V1T10PAQ6)

(d) tan


α 2

tan

β 2


JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 4 of A - 34 Que. 30. Let A, B, C be three angles such that A + B + C = π. If tan A. tan B = csc

π

then the vlaue of

6

is equal to (a)

cosAcosB cosC

(code-V1T12PAQ5)

1

1

(b)

2

(c) 1

3

1

(d)

2

Que. 31. Which value of θ listed below leads to 2sin θ > 1 and 3cos θ < 1 ? (a) 70o

(b) 140 o

(c) 210 o

(code-V1T12PAQ7)

(d) 280 o

Que.32. In a trinalge ABC, a = 3, b = 4 and c = 5. The value of sin A + sin 2B + sin 3C equals (a)

24

14

(b)

25

(c)

25 2

64 25

(d) None.

(code-V1T13PAQ1)

2

Que.33. Let y = ( sin x + cos ecx ) + ( cos x + sec x ) , then the minimum value of y,, ∀ x ∈ R is (a) 7

(b) 8

(c) 9

(code-V1T13PAQ4)

(d) 10

Que.34. If the equation cot 4 x − 2 cosec2 x + a 2 = 0 has atleast one solution then, sum of all possible integral values of ‘a’ is equal to (code-V1T13PAQ5) (a) 4

(b) 3

(c) 2

(d) 0

Que.35. If θ be an acute angle satistying the equation 8 cos 2θ + 8 sec 2θ = 65, then value of cos θ is equal to (a)

1

(b)

8

2

(c)

3

2 3 3

3

(d)

(code-V1T13PAQ7)

4

Que.36. If 2 sin x + 7 cos px = 9 has atleast one solution then p must be

(code-V1T13PAQ8)

(a) an odd integer

(b) an even integer

(c) a rational number

(d) an irrational number

Que.37. If θ∈ ( π / 4, π / 2 ) and

∞

1

∑= tan

n

n 1

(a ) 3

(b)

θ

= sin θ + cos θ then the value of tan θ is (c) 2 + 3

2 +1

(d)

2

Que.38. If sin x + a cos x = b then the value of a sin x − cos x is equal to (a)

a 2 + b2 + 1

(b)

(c)

a 2 + b2 − 1

a 2 − b2 −1

(code-V1T13PAQ11)

(d)

a 2 − b2 + 1

Que.39. The least value of x for 0 < x < π / 2, such that cos(2x) = 3 sin(2x), is (a)

π 12

(b)

2π

(c)

12

3π 12

(d)

(code-V1T13PAQ10)

(code-V1T13PAQ14)

4π 12

Que.40. Let θ∈ [ 0, 4π] satisfying the equation ( sin θ + 2 )( sin θ + 3)( sin θ + 4 ) = 6. if the sum of all value of θ is of the form k π then the value of ‘k’, is (a) 6

(b) 5

(code-V1T13PAQ18)

(c) 4

(d) 2

Que.41. Let f (x) = a sin x + c, where a and c are real numbers and a > 0. Then f (x) < 0 ∀ x ∈ R if (a) c < −a THE “BOND”

(b) c > −a || Phy. by Chitranjan||

(c) − a < c < a

(d) c < a

||Chem. by Pavan Gubrele||

(code-V1T13PAQ19)


JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 5 of A - 34 Que.42. In which one of the following intervals the inequality, sin x < cos x < tan x < cot x can hold good ?

 π (a)  0,   4

 3π  (b)  , π   4 

 5π 3π  (c)  ,   4 2 

 7π  (d)  , 2π   4 

(code-V1T13PAQ21)

Que.43. If sin x + sin y + sin z = 0 = cos x + cos y + cos z then the expression, cos ( θ − x ) + cos ( θ − y )

+ cos ( θ − z ) , for θ =∈ R is

(code-V1T15PAQ1)

(a) independent of θ but dependent on x, y, z (b) dependent on θ but independent of x, y, z (c) dependent on x, y, z and θ (d) independent of x, y, z and θ Que.44. In ∆ABC, AB = 1, BC = 1 and AC = 1 equals (a) 3 2

2. In

(b) 7 4 9

Que.45. The sum

∑= sin n 1

2

nπ 18

(a) 5

∆MNP, MN = 1, and ∠MNP − 2 ∠ABC. The side MP

(c) 2 2

(d)

7 2

equals

(code-V1T15PAQ6)

(b) 4

Que.46. Let area of a triangle ABC is

(code-V1T15PAQ5)

(c) 3 −1 2

(

)

5 +1

, b = 2 and c =

(

(d) 3 5

)

3 − 1 and

∠A is acute. The measure of the

angle C is

(code-V1T15PAQ10)

(a) 15o

(b) 30o

(c) 60o

(d) 75o

2 2 Que.47. If the inequality sin x + a cos x + a > 1 + cos x holds for any x ∈ R then the largest negative intergral value of ‘a’ is (code-V1T18PAQ2)

(a) – 4

(b) – 3

(c) – 2

(d) – 1

 a 1 n sin Que. 48. The value of the expression  2  .  2 + cos ( ka )  , is   k =1 

∑

(a)

1 2

 

sin   n +

1  a  2 

(b)

(c) sin ( ( n + 1) a )

1 2

 

sin   n −

(code-V1T18PAQ3)

1  a  2 

(b) cos n ( a )

Que. 49. Suppose ABC is a triangle with 3 acute angle A,B and C. The point whose coordinates are

( cos B − sin A, sin B − cos A ) can be in the

(code-V1T18PAQ4)

(a) first and 2nd quadrant

(b) second the 3rd quadrant

(c) third and 4th quadrant

(d) second quadrant only

Que. 50. Number of solution of the equation, sin 4 x − cos 2 x sin x + 2 sin 2 x + sin x = 0 is 0 ≤ x ≤ 3π, is (a) 3 THE “BOND”

(b) 4 || Phy. by Chitranjan||

(c) 5

(d) 6


(code-V1T19PAQ3)


JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 6 of A - 34 Que. 51. Let α, β and γ be the angles of a triangle with 0 < α < sin

2

π 2

and 0 < β <

α + sin 2 β = sin γ, then

ο (a) γ ∈ ( 0,30 )

π 2

, satisfying

(code-V1T20PAQ3)

ο o (b) γ ∈ ( 30 , 60 )

ο o (c) γ ∈ ( 60 , 90 )

(d) γ = 90o

Que. 52. Let L and M be the respective intersections of the internal and external angle bisectors of the 2 2 triangle ABC at C and the side AB produced. If CL = CM, then the value of ( a + b ) is (where a & b

have their usual meanings) (a) 2R2

(code-V2T1PAQ4)

(b) 2 2 R 2

(c) 4R 2

(d) 4 2 R 2

Que. 53. In a triangle ABC, if A − B = 120o and R = 8r where R and r have their usual meaning then cos C equals (a) 3/4

(b) 2/3

(c) 5/6

(d) 7/8

(code-V2T3PAQ9)

Que. 54. The system of equations

(code-V2T5PAQ13)

x − y cos θ + z cos 2θ = 0

− x cos θ + y − z cos θ = 0 x cos 2θ − y cos θ + z = 0

. has non trivial solution for

θ equals (b) nπ +

(a) nπ only, n ∈ I. (c) ( 2n − 1)

π 2

only, n ∈ I.

π 4

only, n ∈ I.

(d) all value of θ

a 1 1 1  −1 1 + tan −1 + tan −1 + tan − 1  is expressed as a rational in lowest form ( a + b ) has Que. 55. If tan  tan b 2 3 4 5  the value equal to (code-V2T8PAQ1)

(a) 19

(b) 27

(c) 38

(d) 45

Que. 56. A sector OABO of central angle θ is constructed in a circle with centre O and of radius 6. The radius of the circle that is circumscribed about the triangle OAB, is (code-V2T8PAQ7) (a) 6cos

θ

(b) 6sec

2

θ 2

θ   (c) 3  cos + 2  2  

(d) 3sec

θ 2

Que. 57. Let s, r, R respectively specify the semiperimeter, inradius and circumradius of a triangle ABC. Then ( ab + bc + ca ) in terms of s, r and R given by (a) sr + rR + Rs

(b) R 2 + r 2 + rs

(c) s 2 + R 2 + 2rs

(code-V2T8PAQ8)

(d) r 2 + s 2 + 4Rr

Que. 58. The value of the expression (1 + tan A )(1 + tan B ) when A = 20o and B = 25o reduces to (a) prime number

(b) composite number

(c) irrational number

(d) rational which is not an integer.

THE “BOND”

|| Phy. by Chitranjan||


(code-V2T13PAQ5)


JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 7 of A - 34 cot 2x − tan 2x  π in  0,  equals 8  5π − 8x   1 + sin   2 

Que. 59. The least value of the expression

(a) 1

(b) 2

(c) 4

(code-V2T13PAQ11)

(d) none.

Que. 60. The value of x satisfying the equation sin ( tan −1 x ) = cos ( cos−1 (x + 1) ) is (a)

1

(b) −

2

1

(c)

2

(code-V2T13PAQ22)

(d) No finite value

2 −1

Que. 61. Which one of the following quantities is negative ?

(code-V2T13PAQ23)

(a) cos ( tan −1 (tan 4) ) (b) sin ( cot −1 (cot 4) ) (c) tan ( cos −1 (cos 5) ) (d) cot ( sin −1 (sin 4) ) Que. 62. The product of all real values of x satisfying the equation

 −1  2 − 81 x  2x 2 + 10 x + 4  sin cos  2 cot =  cot   x + 5 x + 3      9x 

 π   + is  2

−1

(a) 9

(b) – 9

(code-V2T13PAQ29)

(c) – 3

(d) – 1

2 2x  2π −1  x − 1  Que. 63. Product of all the solution of the equation tan −1  2 + cot , is  =  x 1 2x 3 −    

(a) 1

(b) –1

(c) 3

(d) − 3

o Que. 64. If the value of tan ( 37.5 ) can be expressed as

( a > b > c > d ) then the value of (a) 1

ad bc

a

− b + c − d where a, b, c, d ∈ N and

is equal to

(b) 2

(code-V2T14PAQ4)

(code-V2T14PAQ13)

(c) 3

(d) 4

Que. 65. If the minimum value of expression y = ( 27 )

cos x

+ ( 81)

sin x

can be expressed in the form

where a, b ∈ N and are in their lowest term then the value of ( a + b ) equals

a/b

(code-V2T17PAQ10)

Que. 66. Let f (x) = sin x + cos x + tan x + arc sin x + arc cos x + arc tan x. If M and m are maximum and minimum values of f(x) then their arithmetic mean is equal to (code-V2T18PAQ3) (a)

π 2

+ cos1

(b)

π 2

+ sin1

(c)

π 4

+ tan 1+ cos1

(d)

π 4

+ tan 1+ sin 1

rπ Que. 67. The sum ∑ cos   equals 6 r =1 2009

(a) 0 3π   Que. 68.  tan  8  

(b) 1 2009

3π   +  − cot  8  

(code-V2T19PAQ1)

(c)

3 +1 2 2

2+ 3 2

2009

is

(code-V2T19PAQ3)

(a) even integer

(b) odd integer

(c) rational which is not an integer

(d) irraional

THE “BOND”

(d)





Comprehesion Type

# 1 Paragraph for Q. 1 to Q. 3 Let r and R denote the radii of the incircle and the circumcircle of the triangle ABC with sides a, b, c and a + b + c = 2s. Also ∆ denotes the area of the trianlge. (code-V1T16PAQ7,8,9) 1.

The value of the product

A

∏ cot 2 equals

cyclic ABC

(a)

2.

r2

(b)

∆

The vlaue of the sum

∆ s

R ( a + b + c)

(c)

2

A

2

(d)

abc

r s

B

∏ cot 2 cot 2 equals

cyclic A,B,C

(a)

3.

R + 4r

(b)

r

4R + r

4R + r

(c)

r

(d)

R

Let f (x) = 0 denotes a cubic whose roots are cot

A 2

, cot

B 2

, cot

C 2

4R r

. If the triangle ABC is such that one

of its anlge is 90o then which one of the following holds good ? (a) r + 2R = s

(b) 3r + 2R = s + 2

(c) 1 + r + 4R = 2s

(d) 4r + R = s

# 2 Paragraph for Q. 4 to Q. 6 Let ABC be an acute triangle with orthocenter H.D,E,F are the feet of the perpendiculars from A,B,and C on the opposite sides. Also R is the circumradius of the triangle ABC. (code-V1T18PAQ6,7,8) 2

2

2

Given ( AH )( BH )( CH ) = 3 and ( AH ) + ( BH ) + ( CH) = 7

4.

The ratio

(a) 5.

∑ cos

2

3

A

(b)

14R

3 7R

(c)

7 3R

(d)

14 3R

The product ( HD )( HE )( HF ) has the value equal to (a)

6.

∏ cosA has the value equal to

9

(b)

64R 3

9 8R 3

(c)

8 9R 3

(d)

9 32R 3

The value of R is (a) 1

THE “BOND”

(b) 31/ 3


(c)

3 2

(d)


5 2



Assertion & Reason Type

In this section each que. contains STATEMENT-1 (Assertion) & STATEMENT-2(Reason).Each question has 4 choices (A), (B), (C) and (D), out of which only one is correct. Bubble (A) STATEMENT-1 is true, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1. Bubble (B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1. Bubble (C)

STATEMENT-1 is True, STATEMENT-2 is False.

Bubble (D)

STATEMENT-1 is False, STATEMENT-2 is True.

Que. 1. Statement - 1 :

(code-V1T4PAQ8)

In a triangle ABC if A is obtuse then tan B tan C > 1 because Statement - 2 : In any triangle ABC, tanA =

tan B + tan C tan B tan C − 1

Que. 2. Statement - 1 : c

(code-V1T4PAQ9) c

cos (10 ) and cos ( −10) both are negative and have the same value.

because Statement - 2 : c

cos θ = cos ( −θ ) and the real numbers (10 ) and

c

( −10 ) both lie in the third quadrant.

Que. 3. Let α, β and γ satisfy 0 < α < β < γ < 2π and cos ( x + α) + cos ( x +β) + cos ( x + γ ) = 0 ∀ x∈ R

γ −α =

Statement - 1 :

2π 3

.

(code-V1T6PAQ1)

because cos α + cos β + cos γ = 0 and sin α + sin β + sin γ = 0.

Statement - 2 : Que. 4. If A + B + C = π then Statement - 1 :

(code-V1T6PAQ3)

cos 2 A + cos 2 B + cos 2 C has its minimum value

3 4

.

because Statement - 2 :

1

Maximum value of cos Acos Bcos C = . 8

Que. 5. Let f (x) = 3sin 2 x + 4sin xcos x + 4 cos 2 x, x ∈ R. Statement - 1 :

Greatest and least values of f (x) ∀ x ∈ R are

(code-V1T8PAQ11)

7 + 17 2

and

7 − 17 2

respectively..

because Statemtnt 2 :

a + b − a 2 + b 2 + c 2 − 2ab 2

≤ a sin 2 x + b sin x cos x + c cos 2 x ≤

a + b + a 2 + b2

+ c 2 − 2ab

2

where a, b, c ∈ R

THE “BOND”




JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 10 of A - 34 tan

Que. 6. Statement - 1 :

6π 7

− tan

5π 7

− tan

π 7

= tan

6π 7

.tan

5π 7

. tan

π

(code-V1T10PAQ7)

7

because If θ = α + β, then tan θ− tan α− tan β = tan θ.tan α.tan β.

Statement - 2 :

Que. 7. Consider the following statements Statement - 1 :

(code-V1T12PAQ8)

In any right angled triangle ABC, sin 2 A + sin 2 B + sin 2 C = 2

because 2 2 2 In any triangle ABC, sin A + sin B + sin C = 2 − 2 cos A cos B cos C ]

Statement - 2 : Que. 8. Statement-1:

= n cot

A 2

+ n cot

In B 2

any

+ n cot

triangle

 

 n  co t

ABC

A 2

+ co t

B 2

+ co t

C   2 

C 2

because

(code-V1T14PAQ5)

Statement - 2 : Que. 9. Statement - 1 :

(

(

n 1 + 3 + 2 + 3

) ) = n1 + n

General solution of

(

3 + n 2 + 3

tan 4x − tan 2x 1 + tan 4x tan 2x

= 1 is x =

)

nπ 2

because

π + ,n∈I 8

(code-V1T18PAQ11)

Statement - 2 : Que. 10. Statement - 1 :

π

General solution of tan α = 1 is α = nπ + , n ∈ I. 4

The equation sin ( cos x ) = cos (sin x ) has no real solution (code-V1T19PAQ10)

because Statement - 2 : Que. 11. Statement 1:

sin x ± cos x is bounded in  − 2, 2 

In any triangle ABC, cot A + cot B + cot C > 0

(code-V2T7PAQ10)

because Statement 2: Minimum value of cot A + cot B + cot C in any triangle ABC is 1. Que. 12. Let P be the point lying inside the acute triangle ABC such that angles sutended by each side at P is 120o. Equilateral triangles AFB, BDC,CEA are constructed outwardly on sides AB, BC, CA of

∆ABC Statement 1: Lines AD, BE, CF are concurrent at P.

(code-V2T18PAQ8)

because Statement 2: P is the radical centre of circumcircles of triangles ABF, BDE, CEA Que. 13. Let ABC be an acute triangle whose orthocentre is at H. Altitude from A is produced to meet the circumcircle of the triangle ABC at D. (code-V2T19PAQ7) Statement 1: The distange HD = 4R cosBcosC where R is the circumradius of the triangle ABC. because Statement 2: Image of orthocentre H in any side of an acute triangle lies on its circumcircle.

THE “BOND”





More than One May Correct Type

Que. 1. If 2cos θ + 2 2 = 3sec θ where θ∈ ( 0, 2π ) then which of the following can be correct ? 1

(a) cos θ =

2

(b) tan θ = 1

(c) sin θ = −

1

(d) cot θ = −1

2

(code-V1T2PAQ9)

Que. 2 Which of the following real numbers when simplified are neither terminating nor repeating decimal ? (a) sin 75o.cos75o

(b) log 2 28

(d) 8−(log27 3)

(c) log3 5.log5 6

(code-V1T2PAQ10)

Que. 3. Suppose ABCD (in order) is a quadrilateral inscribed in a circle. Which of the following is/are always True ?

(code-V1T2PAQ11)

(a) sec B = sec D

(b) cot A + cos C = 0

(c) cos ecA = cos ecC

(d) tan B + tan D = 0

Que. 4. which of the following quantities are rational ?

(code-V1T4PAQ15)

 11π   5π  (a) sin   sin    12   12 

 9π   4π  (b) cos ec   sec    10   5 

4 π 4 π (c) sin   + cos   8 8

2 π  4π  8π   (d) 1 + cos  1 + cos  1 + cos  9  9  9  

Que. 5. In a triangle ABC, a semicircle is inscribed, whose diameter lies on the side c. If x is the length of the angle bisector through angle C then the radius of the semicircle is (code-V1T6PAQ6) (a)

abc

(b)

4R (sin A + sin B) 2

(c) xsin

C

(d)

2

∆ x 2 s(s − a)(s − b)(s − c) s

Que. 6. The possible value(s) of x satisfying the equation sin x − 3sin 2x + sin 3x = cos x − 3cos 2x + cos 3x, is/ (code-V1T6PAQ9) are. (a) −

7π 8

(b) −

π 8

(c)

π 8

Que. 7. In which of the following sets the inequality sin 6 x + cos6 x >

 π π (a)  − ,   8 8

 3π 5π  , (b)    8 8 

(d)

 π 3π  (c)  ,  4 4 

5 8

π 4

holds good ?

 7π 9π  (d)  ,   8 8 

Que. 8. The value of x satisfying the equation cos ( n x ) = 0, is (a) eπ / 2 THE “BOND”

(b) e−π / 2 || Phy. by Chitranjan||

(c) eπ

(code-V1T6PAQ12)

(code-V1T10PAQ9)

(d) e3π / 2



JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 12 of A - 34 Que. 9. Which of the following do/does not reduce to unity ?

(code-V1T10PAQ11)

sin (180o

+ A ) cot ( 90o + A ) cos ( 360o − A ) cos ecA . . (a) sin ( − A ) tan (180o + A ) tan ( 90o + A )

(b)

sin ( −A ) tan (180o + A )

−

tan ( 90o + A ) cot A

+

cosA sin ( 90o + A )

− sin 6o sin 66o (c) sin 21o cos 39 o − cos 51o sin 69 o sin 24o cos 6o

(d)

cos ( 90o + A ) sec ( −A ) tan (180o − A ) sec ( 360o

+ A ) sin (180o + A ) cot (90o − A )

Que. 10. Which of the following identities wherever difined hold(s) good ?

(code-V1T10PAQ12)

(a) cot α − tan α = 2 cot 2α

(b) tan ( 45o + α ) − tan ( 45o − α ) = 2 cos ec2 α

(c) tan ( 45o + α ) + tan ( 45o − α ) = 2sec2 α

(d) tan α + cot α = 2 tan 2α.

Que. 11. Let α, β and γ are some angles in the1st quadrant satisfying tan ( α + β ) =

15 8

which of the following holds good ?

and cos ecγ =

17 then 8

(code-V1T14PAQ7)

(a) α + β + γ = π (b) cot α.cot β.cot γ = cot α + cot β + cot γ (c) tan α + tan β + tan γ = tan α.tan β.tan γ (d) tan α. tan β + tan β.tan γ + tan γ. tan α = 1 Que. 12. Which of the following statements are always correct ? (where Q denotes the set of rationals) (a) cos 2θ ∈ Q and sin 2θ∈ Q

⇒

(b) tan θ∈ Q

⇒

tan θ ∈ Q (in difinde)

(code-V1T14PAQ8)

sin 2θ, cos 2θ and tan 2θ∈ Q (if defined)

(c) If sin θ∈ Q and cos θ∈ Q (d) if sin θ∈ Q

⇒

⇒

tan 3θ∈ Q (if defined)

cos 3θ∈ Q

Que. 13. Given that sin 3θ = sin 3α, then which of the following angles will be equal to cos θ ?

π  (a) cos  + α  3 

π  (b) cos  − α  3 

 2π + α  (c) cos   3 

 2π − α  (d) cos   3 

Que. 14. If the quadratic equation ( cos ec2 θ − 4 ) x 2 + ( cot θ + 3 ) x + cos2

3π 2

= 0 holds true for all real x then

the most general values of θ can be given by (a) 2nπ +

11π 6

THE “BOND”

(b) 2nπ +

5π 6


(c) 2nπ ±

(code-V1T14PAQ9)

(code-V1T15PAQ12)

7π 6

(d) nπ ±


11π 6


JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 13 of A - 34 Que. 15. If α and β are two different solution of a cos θ + b sin θ = c, then which of the following hold(s) good ? (code-V1T15PAQ13) (a) sin α + sin β = (c) cos α cos β =

2bc a2

(b) sin α sin β =

+ b2

2ac a2

c2 − a 2 a 2 + b2

(d) cos α cos β =

+ b2

c2 + a 2 a 2 + b2

Que. 16. If in a triangle ABC, cos 3A + cos 3B + cos 3C = 1, then

(code-V1T15PAQ14)

(a) ∆ABC is a right angled triangle. (b) ∆ABC is an obtuse angle triangle (c) ∆ABC is an acute angled triangle (d) Inradius ‘r’ of the triangle ABC is either

3 ( s − a ) or

Que. 17. The value of x in (0, π / 2) satisfying the equation, (a)

π

(b)

12

5π

(c)

12

3 ( s − b ) or

3 −1 sin x

+

7π

3 +1 cos x

= 4 2 is

(d)

24

3 ( s − c) . (code-V2T2PAQ11)

11π 36

Que. 18. In a triangle ABC, 3sin A + 4cos B = 6 and 3cos A + 4sin B = 1 then ∠C can be (code-V2T8PAQ12) (a) 30o

(b) 60 o

(c) 90o

(d) 150 o

Que. 19. If cos 3θ = cos 3α then the value of sin θ can be given by

π  (b) sin  ± α  3 

(a) ± sin α

 2π  + α (c) sin   3 

(code-V2T15PAQ9)

 2π  − α (d) sin   3 

Que. 20. Let ABC be a triangle with AA1, BB1, CC1 as their medians and G be the centriod. If the points A, C1, G, B1 be concyclic then which one of the following relations do/does not hold good ? (code-V2T19PAQ10) (a) 2a 2 = b 2 + c 2

(b) 3b 2 + c 2 + a 2

(c) 4c 2 + a 2 + b 2

(d) 3a 2 + b 2 + c 2

Match Matrix Type Que. 1. Match the general solution of the trigonometric equation given in Column - I with their corresponding entries given Column - II. (code-V1T14PBQ1) Column - I

Column - II

π

π ∪ nπ+ , n ∈ I.

cos 2x + cos x = 1

P.

x = nπ+

B.

cos x = 3 (1 − sin x )

Q.

x

=

C.

1 + 3 tan 2 x = 1 + 3 tan x

R.

x

π = ( 2n − 1) , n ∈ I.

D.

tan 3x − tan 2x − tan x = 0

S.

x = 2nπ +

A.

2

2

(

THE “BOND”

)

Subjective ||Chem. Type ( by Up to 4 Gubrele|| digit) || Phy. by Chitranjan|| Pavan

nπ 3

4

6

,n ∈I

6

π 2

π ∪ 2nπ+ ,n ∈I. 6


JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 14 of A - 34 Que. 1. The expression 2 cos

π 17

.cos

9π 17

+ cos

7π 17

+ cos

9π 17

simplifies to an integer P. Find the value of P..

(code-V1T1PAQ1)

sin ( α + β ) − 2sin α + sin ( α − β)

Que. 2. Show that the expression cos α + β − 2 cos α + cos α − β is independent of β. (code-V1T1PAQ2) ( ) ( ) sin θ.sin 2θ + sin 3θ sin 6θ + sin 4θ sin13θ

Que. 3. If the expression k.

sin θ cos 2θ + sin 3θ cos 6θ + sin 4θ cos13θ

= tan kθ, where k ∈ N. Find the value of

(code-V1T1PAQ4)

8 Que. 4. If y = cos

x 2

x π − sin 8 . Find the value of y when x = and also when x = π . 2

4

(code-V1T3PAQ4)

6

Que. 5. In a trianlge ABC, given sin A : sin B : sin C = 4 : 5 : 6 and cos A : cos B : cos C = x : y : z. If the ordered 2 2 2 pair (x,y) satisfies this, then compute the value of ( x + y + z ) where x, y, z ∈ N and are in their

lowest form. Que. 6. Let

(code-V1T7PBQ2)

A = cos 360o.sin 2 270o − 2 cos180o . tan 225o ,

C = 2 sec 2 2 π.cos 0c

+ 3sin 3

3π 2

− cos ec

5π 2

B = 3sin 540o.sec 720o

and D = tan π.cos

3π 2

+ 2 cos ec450o − cos 3600o 3π

+ sec 2π − cos ec

2

A + B − C ÷ D.

. Find the value of

(code-V1T9PAQ1)

Que. 7. If the value of the expression E = cos 4 x − k 2 cos 2 2x + sin 4 x, is independent of x then find the set of (code-V1T9PAQ3) values of k. Que. 8. If cot

π 24

= p + q + r + s where p, q, r, s ∈ N, find the value of ( p + q + r + s ) .

Que. 9. Let L denotes the value of the expression, tanxtan2x

sin 2θ − sin 6θ + cos 2θ − cos 6θ sin 4θ − cos 4θ

and

M denotes the value of

and

N denotes the numerical value of the expression (wherever difined)

tan 2x − tan x

(code-V1T9PAQ5)

, when θ = 27 o

when x = 9o.

(code-V1T9PAQ7)

1 − cos 4α sec 2α − 1 2

+

1 + cos 4 α cos ec 2 2α − 1

when it is simplified. Find the value of the product (LMN). 4

5

5

13

Que. 10. If cos ( α + β ) = ;sin ( α − β) =

and α, β lie between 0 and

π 4

, then find the value of tan 2β.

(code-V1T11PAQ1)

Que. 11. Find the range of values of k for which the equation 2 cos4 x − sin 4 x + k = 0, has atleast one solution. (code-V1T11PAQ3) 2π  4π 3  + sin 3  θ +  − = − sin 3θ. Que. 12. Prove that, sin 3 θ + sin 3  θ +  3  3  4  

(code-V1T11PAQ5)

 tan2r −1  Que. 13. Compute the value of the sum ∑  r  r =1  cos2 

(code-V1T11PAQ6)

n

THE “BOND”




JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 15 of A - 34 Que. 14. If 15sin 4 α + 10 cos4 α = 6 then find the value of 8 cosec6 α + 27 sec6 α.

(code-V1T15PDQ1)

Que. 15. Given that x + sin y = 2008 and x + 2008cos y = 2007 where 0 ≤ y ≤ π / 2. Find the value [ x + y ]. (Here [x] denotes greatest interger function)

(code-V2T1PDQ2)

φ2 (θn ) where Que. 16. The sum ∑ 2 sin x .sin 1[1 + sec(x − 1).sec(x + 1)] can be written in the form as ∑ (−1) ψ (θ n ) n =1 x =2 4

44

n

φ and ψ are trigonometric functions and θ1 , θ2 , θ3 , θ 4 are in degrees ∈ [0,45]. Find ( θ1 + θ2 + θ3 + θ 4 ) . (code-V2T17PDQ1)

Que. 17. If the total between the curves f (x) = cos −1 (sin x) and g(x) = sin −1 (cos x) on the interval [ −7 π, 7 π] is A, find the value of 49A. (Take π = 22/7 ) (code-V2T17PDQ3)

[SOLUTION] Single Correct Type Que. 1. (D) cos A cos C + ( − cos C )( − cos A ) cos A sin C − ( sin C )( − cos A )

Que. 2 (B) 3a; = 4;a = log 3 4; Hence abcdef

2cosAcosC

⇒

2 cos A sin C

= + cosC

b = log 4 etc.

|||1y

= log3 4; log4 5; log5 6.log6 7.log7 8.log8 9 = log3 9 = 2

⇒

2.

Que. 3. (D) cos15o

= 2 + 3 ≅ 3.732;

cos ec15o

=

4 6− 2

tan 60o

= 3 ≅ 1.732; sec15o =

4 6+ 2

= 6 − 2 = 1.035;

= 6 + 2 = 3.86 which is largest

Que. 4. (C)

α−γ   α + γ  cos    sin α − sin γ  2   2  = cot  α + γ  = cot = β   cos γ − cos α 2  α − γ   α + γ    2 sin  sin     2   2  2sin 

2 Que. 5. (B) 1 + 1 + tan 2A = 1 + sec 2A

tan 2A

1 − sec 2A 1 − cos 2A  = −  ( 2A = 220 ) =  = − tan A. o

tan 2A

tan 2A

 sin 2A 

2 2 2 2 2 Que. 6. (C) y = 256 ( sin x + cos ec x ) + 68cos ec x, 256 ( (sin x − cos ecx) + 2 ) + 68cos ec x

Minimum when x = Que. 7. (a)

E=

THE “BOND”

−1 + sec A tan A

=

π 2

or

1 − cos A sin A

π − and minimum vlue = 512 + 68 = 580 2

= tan


A 2



JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 16 of A - 34 Que. 8. (B).

Que. 9. (A) Let R ' be the radius of the circumcircle of triangle ABC using sine law in trianlge BPC a sin120 a

o

= 2R1

................. (1)

also

= 2R sin 60o ( R = 1 given ) a = 3;

Que. 10. (D) ∆ =

1 2

ab ⇒ ab = 60

c=

a o

sin60

R1 =

form (1)

a 2 + b2

= 2R ( in ∆ABC ) 2 3 1 . = 1. 3 2

also a 2 + b 2 ≥ 2ab B

∴ ∴

a 2 + b2

≥ 2ab

A

equality occurs when a = b

C

minimum value of Alternatively: b = c cos θ; c2 |min = 120

Que. 11. (D) In (D) it should be Que 12. (A) put

a

a

2

5 −1 8

B

Θ A

+ b2 = 2 ab = 120 = 2 30

a = c sin θ

⇒

C

1

c 2 sin 2θ

2

4

∆ = c2 sin θ cos θ =

= 3θ

c

2

= 120 cos ec2θ

c = 2 30

.

= 2R sin A etc.

T1 = 2R sin ( B + C ) cos ( B − C ) = R [sin 2B + sin 2C ]etc. E = R [sin 2B + sin 2C + sin 2C + sin 2A + sin 2A + sin 2B] = 2R ( sin 2A + sin 2B + sin 2C )

= 8R.

a

.

b

.

c

2R 2R 2R

=

abc R2

.

Que. 13. (D) where n ∈ I. sin 2 2x = 2 cos2 x 4sin 2 x cos2 x = 2 cos2 x cos2 x 1 − 2 sin2 x

cos 2 x = 0

b

Que. 14. (B) c

sin 2 x

or

=

3 +1 b − c ; 2 b+c

now using tan

THE “BOND”

B−C 2

=

=

b−c b+c

=

1

∴

2

3 + 1− 2 3 + 1+ 2

cot

A 2

=

=

x 3 −1

(

)

3 +1

3 −1

(


= nπ ±

)

3 +1

.

.

3 3

π 2

=0

or

x

= nπ ±

π 4

1 3

= 2− 3 ⇒

B− C 2

= 15o ∴ B − C = 30o



JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 17 of A - 34 Que. 15. (C) Que. 16. (D) sin 2 θ = 1 [sin θ ≠ +1] ⇒ sin θ = −1 ⇒ θ = 2nπ − π / 2 ⇒ infinite roots

⇒

Que. 17. (C) In a triangle b + c > a

∴ a ( b + c − a ) > 0 |||1y

b ( c + a − b) > 0

c ( a + b − c) > 0

and

a ( b + c) + b ( c + a ) + c ( a + b ) > a

∴

b+c−a > 0

a2 + b2 + c2 ab + bc + ca

∴

<2

+ b2 + c2

2 ( ab + bc + ca ) > a

............(1) also forany a,b,c

+ b2 + c2

∈ R a 2 + b 2 + c 2 ≥ ab + bc + ca

∑a < 2 1≤ ∑ ab 2

........(2) (equality holds if a = b = c)

Que. 18. (A) Make a quadratic in cos2θ to get cos 2α + cos 2β =

⇒

2

2

+ b2 + c2 ≥1 ab + bc + ca a

2

2 ( cos

2

α + cos 2 β ) =

2ac a

2

+b

+ 2; 2

cos

2

form (1) and (2)

2ac a

α + cos2 β =

2

+ b2

a 2 + ac + b2 a

2

+ b2

 b 2  2b 2 a 1 − 2  + a (1 − tan 2 θ ) 2b.tan θ a ( a 2 − b 2 ) + 2ab 2 a ( a 2 + b 2 ) a  a  = = 2 2 = a. Que. 19. (A) 1 + tan 2 θ + 1 + tan2 θ = b2 a 2 + b2 a +b 1+

Que. 20. (D) f 4 ( x ) − f 6 ( x ) = 1

1

1 sin x + cos x ) − (sin ( 4 6 4

2

 1

1

a2



3

1

6

1

x + cos x ) =

(1 − 2sin 4

6

1

2

cos x ) − 2

1

(1 −3sin 6

2

cos x ) 2

1

= 1 − sin 2 2x  − 1 − sin 2 2x  = − = . 4 2  6 4  4 6 12 Que. 21. (A) Que. 22. (D) Given A = 60o ; tan 30o =

cosB =

1 − tan 2 ( B 2 ) 1 + tan 2 ( B 2 )

=

r 5

5

⇒r=

3

1 − ( 25 27 ) 1 + ( 25 27 )

=

now tan

2 521

=

1 26

B 2

r

5

3

3 3

= =

(a = ?)

26

=

675 15 3

B 1 C

sin B =

Hence

3 1 1 15 3 + . . 2 26 2 26

=

15 3 26

b sin C = sin ( A + B )

1

A

30o 30o

5

16 3  = 4 3 = sin C ⇒ c = a ;  52  13 sin C sin A

a I r D B/2 c 3 a=

= sin A cos B + cos A s in B = B

8. 3 13 . 2 4 3

= 13.

Que. 23. (B) THE “BOND”




JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 18 of A - 34 Que. 24. (A,B,D)

(c)

∑ sin

2

A

1

3

1

2

2

2

= 3 − ( cos A + cos B + cos C )  = − ( cos A + cos B + cos C )

2

but [ cos A + cos B + cos C ]max =

3 2

∑ sin

∴

2

A

3

3

3

= − = ∴ 2  min 2 4 4

∑ sin

2

A 2

3

≥ ⇒ (c) si wrong. 4

a,b,c are correct and hold good in an equilateral triangle as their maximumum values. (b) RHS =

Que. 25. (A,B,C)

sin 4α + sin 2α sin 2α.sin α 4

=

2sin 3α.cos α

=

sin 2α.sin 4α

1 sin α

= cos ec α ( using π = 7 α) ⇒ (b).

(c)

cos α + cos 3α + cos 5α sum of a series with constant d = 2α sum =

(d)

continued product = 1 ⇒ (d) is also wrong. (a) x = π 8, ⇒ ( tan x )

Que. 26. (A,B,C) (b)

x = π 6, ⇒ 4n 2

(c)

x = π 2, 2n 2

(d)

x = π 4, 20

Que. 27. (B)

( cos

2

n ( sin x )

> 1 and ( cot x )

n (sin x )

⇒ (c) is wrong.

2

< 1 ⇒ True.

< 5n 2 ⇒ True.

< 3n 2 ⇒ True.

> 2− n 2 ⇒ 1 <

1 2

is not correct ⇒ False.

n 2

A − sin 2 B ) − ( sin 2 A − sin 2 B ) = cos2 A − sin 2 A

(C)

1

= cos 2A

(D) − sin 2B

sin 2B;

2 2 2 2 Que. 28. (D) (1 − sin 5x ) − (1 − sin x ) + sin 4x sin 6x = 0 ⇒ sin x − sin 5x + sin 4x.sin 6x = 0

− sin 6x.sin 4x + sin 4 x.sin 6x = 0 which is true for all x ∈ [ 0, π] ⇒ it is indentity  α +β   α −β  α+β  α +β  α −β  α +β   cos   + 2 sin   cos   cos   + cos    2   2   2   2 =  2   2  Que. 29. (A)  α +β   α −β  α +β  α+β   α −β  α +β  2sin   cos   − 2 sin   cos   cos   − cos    2   2   2   2   2   2  2 sin 

2 cos

= 2sin

α 2

α 2

cos sin

β 2

β

α β = cot cot . 2

2

2

Que. 30. C. Given tan A. tan B = 2 Let y =

cos A cos B cos C

=−

cos A.cos B cos(A + B)

=

cos A.cos B sin A sin B − cos A cos B

=

1 tan A tan B − 1

=

1 2 −1

=1

Que. 31. B. 2sin θ > 1 ⇒ sin θ > 0 ⇒ θ∈1st or 2 nd quadrant , 3sin θ < 1 ⇒ cos θ < 0 ⇒ θ∈ 2nd or 3rd quadrant hence

θ∈ 2nd ⇒ possible answer is (B). A Que. 32. B.

E = sin A + sin 2B + sin 3C ⇒ E =

3 5

4 3

15

5 5

25

+ 2. . −1 =

+

24 25

−1 =

39 − 25 25

=

14 25

5

.

B THE “BOND”



3

4 C


JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 19 of A - 34 y = ( sin 2 x + cos2 x ) + 2 ( sin x cos ecx + cos x sec x ) + sec2 x + cos ec2 x

Que. 33. C.

2

= 5 + 2 + tan 2 x + cot 2 x = 7 + ( tan x − cot x ) + 2 ∴ ymin = 9. cot 4 x − 2 (1 + cot 2 x ) + a 2

Que.34. D.

2

= 0 ⇒ cot 4 x − 2 cot 2 x + a 2 − 2 = 0 ⇒ ( cot 2 x − 1) = 3 − a 2 to have

atleast one solution 3 − a 2 ≥ 0 ⇒ a 2 − 3 ≤ 0 ⇒ a ∈  − 3, 3  integral values −1, 0,1 ∴ sum = 0. Que.35. D.

cos 2θ = t ∴ 8t +

Let

8 t

= 65 ⇒ 8t 2 − 65t + 8 = 0 ⇒ 8t − 64t − t + 8 ⇒ 8t ( t − 8) − ( t − 8) = 0

1

1

1

9

8

8

8

16

⇒ t = 8 or t = (t = 8 is rejected, think !) ∴ cos 2θ = ; 2 cos 2 θ − 1 = ⇒ cos2 θ = 2sin x + 7 cos px

Que.36. C.

x = (4n + 1)

π 2

1 tan θ

2mπ

π 2mπ 4m ⇒p= ( m, n,∈ I ) ∴ (4n + 1) = ∴ p ∈ rational. p 2 p 4n + 1

<1

1

now,

tan θ

1

⇒

tan θ 1

1−

= sin θ + cos θ ⇒

1 tan θ − 1

= sin θ + cos θ ⇒

Que.38. D.

1 2

=

π 3

a 2 + 1 = k 2 + b2 ⇒ k2

cos 2x 2

− π 3

3

cos θ sin θ − cos θ

1

+

θ tan 2 θ

+ ........∞ = sin θ + cos θ

= sin θ + cos θ

⇒ tan θ = 3.

2

= a 2 − b2 + 1 ⇒ k = a 2 − b2 + 1.

sin 2x = 0 or cos 2x.cos 2x.cos

π 3

π π  − sin 2x.son ⇒ cos  2x +  = 0 3



3

π π π = ; 2x = ⇒ x = . 2

6

12

( sin θ + 2 )(sin θ + 3)(sin θ + 4 ) = 6 LHS ≥ 6 and RHS = 6 ⇒ equality only can hold if

sin θ = −1. ⇒ sin θ = −1 ⇒ θ =

Que.41. A.

tan

2

Let a sin x − cos x = k, k ≥ 0 ............... (1) also sin x + a cos x = b ............... (2) Square and

⇒ 2x + Que.40. B.

1

⇒ 2 cos2 θ + cos θ − 1 = 0 ⇒ ( 2 cos θ − 1)( cos θ + 1) = 0

or cos θ = −1 (rejected) ⇒

add (1) and (2)

Que.39. A.

+

tan θ

⇒ cos θ = sin 2 θ − cos 2 θ = 1 − 2 cos 2 θ cos θ =

4

= 9 is possible only if sin x = 1 cos px = 1

and px = 2mπ ⇒ x =

tan θ > 1 ⇒ 0 <

Que.37. A.

3

⇒ cos θ = .

3π 7 π , 2 2

c a sin x + c < 0 ⇒ sin x < − ; a

THE “BOND”

∴ sum = 5 π ⇒ 5. c

c

a

a

− > sin x; − > 1; − c > a ⇒ a + c < 0 ⇒ (A)




JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 20 of A - 34 Que.42. A.

In 2nd quadrant sin x < cos x is False (think !) In 4th quadrant cos x < tan x is False (think !)

 5π 3π  in 3rd quadrant, i.e.  ,  if tan x < cot x ⇒ tan 2 x < 1 which is not correct hence A can be correct  4 2 

π



now

sin x < cos x is true in  0, 4  and tan x < cot x is also true  

∴

only the value of x for which cos x < tan x is be determined

∴

now

⇒ sin x =

2 cos x = tan x i.e. cos x

−1 ± 5 2



2



Que.43. D.

cos θ (

,

π

 −1 5 − 1   and cos x > tan x in  0, sin  4 4  

∑ cos x ) + sin θ ( ∑ sin x) = 0 1

1+1−

Que.44. D.

2

5 −1

∴ cos x < tan x in  sin −1

 5 −1  ⇒ x = sin −1    2 

5 −1

=

; sin x

= sin x or 1 − sin 2 x = sin x ⇒ sin 2 x + sin x − 1 > 0

cos θ =

2

2

agin x 2

=

3

∴ cos 32θ = 2 cos 2 θ − 1 = 2.

4



B

sum = sin 2

Que.46. A. Using ∆ =

Que.47. B.

∴

B−C 2

=

π 18 1 2

b−c b+c

0

THE “BOND”

16

=

1 8

+ sin 2

2π 18

bc sin A ∴

cot

A 2

1

=

8

8

4

2

N

+ ........ + sin 2 1 2

.2

x

1

C

1

sin 2 x + a cos x + a 2

−2

2

M 1/ 2

1

⇒ tan

16

−1=

1 7.2 7 7 = 1 + 1 − 2cos 2θ = 2 (1 − cos 2θ ) = 2 1 −  = = ⇒x= A

Que.45. A.

9

(

3− 3 3 +1

)

20 1

8π 18

+1 now

3 − 1 sin A = 3 +1

.

3 −1

P

3 −1 2

sin 2

π 18

+ sin 2

8π 18

= 1 etc. ⇒ sum = 5.

1

∴ sin A = ⇒ A = 30o 2

= 3 ⇒ B − C = 120o also B + C = 150o ⇒ C = 15o.

2 2 > 1 + cos x put x = 0 ⇒ a + a > 2 ⇒ a + a = 2 > 0 ⇒ ( a + 2 )( a − 1) > 0

∴


largest negative integral value of ‘a’ = – 3. ||Chem. by Pavan Gubrele||



 a 1 n a  1  sin   .  + ∑ cos(ka)  ⇒ sin   .  + cos a + cos 2a + cos 3a + ...... + cos na   2   2 k =1 2 2  

Que. 48. A. 1 2 1

sin

sin

1 

a

a 

3a



1



1  

1

a

    + sin  n +  a − sin  = sin  n +  a. 2 2 2 2 2 2 a

1

1

1

    Que. 49. D. Since ABC are acute angle 2



3a 

5a

+  sin − sin  +  sin − sin  + ........ +  sin  n +  a − sin  n −  a   2 2  2 2  2 2  2 2     

π

∴ A + B > π/ 2 ⇒ A > B>

Again,

π 2

2





− B ⇒ sin A − cos B > 0 ⇒ cos B − sin A < 0

................. (1)

− A ⇒ sin B > cos A ⇒ sin B − cos A > 0

................. (2)

Form (1) and (2) x-coordinate is – ve and y-coordinate is +ve line in 2nd quadrant only.. ⇒ Que. 50. B.

sin x − cos sin x + 2sin x + sin x 4

2

⇒ sin x sin 3 x − cos2 x + 2 sin x + 1 = 0

=0

2

⇒ sin x sin 3 x − 1 + sin 2 x + 2 sin x + 1 = 0

⇒ sin x sin 3 x + sin 2 x + 2 sin x  = 0

⇒ sin 2 x = 0 or sin 2 x + sin x + 2 = 0

⇒ not possible for real x. sin x = 0

⇒ x = 0, π, 2 π, 3 π, Que. 51. D. Que. 52. C.

a 2 + b2

Que. 53. D.

R

⇒

4 solution.

= 4R 2 sin 2 ( 45o − θ ) + sin 2 (135o − θ)  = 4R 2 sin 2 ( 45 o − θ) + cos 2 ( 45 o − θ)  = 4R 2.  

= 8r = 8  4R sin

A 2

sin

B 2

sin

C

∴ 2 sin

 2

A−B A+B C 1  ⇒  cos − cos  sin = 2 2  2 16  sin

2

C 2

1

C

2

2

− sin

cos C = 1 − 2 sin 2

+

1 16

C 2

⇒

A 2

sin

sin

B 2

sin

C 2

=

1 16

C 1 C 1 .  − sin  = ; 2 2 2  16

2

⇒

=0 1

7

8

8

C 1 − sin 4  =0 2 

⇒

sin

C 2

=

1 4

=1 − = .

− cos θ cos 2θ − cos θ − cos θ = 0 using C1 → C1 → C3 1 cos 2θ − cos θ 1 1

Que. 54. D.

For non trivial solution

2 sin 2 θ

1 − cos θ cos 2θ − cos θ cos 2θ 2 0 1 1 − cos θ = 0 ⇒ 2 sin θ 0 − cos θ = 0 ⇒ sin 2 θ = 0 1 1 −2 sin 2 θ − cos θ −1 − cos θ

or

1 1 − cos 2 θ  − 1  cos2 θ − cos 2θ  ⇒ sin 2 θ −  cos2 θ − (cos2 θ − sin2 θ  ⇒ sin2 θ − sin2 θ = 0

hence THE “BOND”

D=0

∀θ∈ R ⇒


D.



JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 22 of A - 34 9 1 1 + + 1  −1 4 5  ; α = tan −1  9  = 19 = 28 = 14 = a ⇒ 14 + 5 = 19. π  α = tan     tan  + α  when  19  1 − 9 10 5 b 1− 1  4   20  19

Que. 55. A.

=

R

Que. 56. D.

abc 4∆

;

1 θ c θ ⇒ c = 12 sin . ∆ = .6.6 sin θ = 18sin θ ⇒ a = b = 6 ⇒ sin = 2

2

12

2

Alternatively :

OD 2

= 62 + x 2 ⇒

x 6

= tan ( θ / 2 ) ⇒ OD2 = 62.sec2 ( θ / 2 ) ⇒ 4r 2 = 36 sec 2 ( θ / 2 ) ⇒ r = 3sec ( θ / 2 ) .

∆ 2 = s ( s − a ) ( s − b )( s − c ) ⇒ r 2s = ( s − a )(s − b )(s − c )

Que. 57. D.

(∴

r = ∆ / s)

= s3 − s 2 ( a + b + c ) + ( ab + bc + ca ) s − abc ∴ r 2s = s3 − 2s 3 + ( ab + bc + ca ) s − abc using

abc 4R

= ∆ ⇒ abc = 4Rrs ⇒

r2

= ( ab + bc + ca ) − s 2 − 4Rr ∴

∑ ab = r

2

+ s 2 + 4Rr.

E = (1 + tan A)(1 + tan B) = 1 + tan A + tan B + tan A tan B

Que. 58. A.

tan(A + B) =

Now

tan A + tan B 1 − tan A tan B

= 1 ⇒ A = 20o and B = 25o ⇒ 1 − tan A tan B = tan A + tan B

⇒ tan A + tan B + tan A tan B = 1 Que. 59. B.

∴

E = 2. ⇒ A.

Expression reduces to 2 corosec 8x

Que. 60. D. x 1+ x2

x2

x +1

=

2

( x + 1) + 1

2

⇒ x 2  (x + 1)2 + 1 = ( x + 1)  (x 2 + 1) ⇒ x 2 (x + 1)2 + x 2 = x2 (x + 1)2 + (x + 1)2

= (x + 1) 2 ⇒ x + 1 = x not possible as x → ∞ or x + 1 = − x ⇒ x = −1/ 2 not possible (think!).

Que. 61. D.

(A)

cos ( tra −1 ( tan(4 − π) ) ) = cos ( 4 − π ) = cos ( π − 4 ) = − cos 4 > 0

(B)

sin ( cot

(C)

tan (cos−1 ( cos ( 2 π − 5 ) ) ) = tan ( 2π − 5 ) = − tan 5 > 0

( as tan 5 < 0 )

(D)

cot ( sin −1 ( sin ( π − 4 ) ) ) = cot ( π − 4 ) = − cot 4 < 0

⇒

THE “BOND”

−1

( cot ( 4 − π ) ) ) = sin ( 4 − π ) = − sin 4 > 0


(as sin 4 < 0)

(D) si correct.



JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 23 of A - 34 Que. 62. A.

  2  2 ( x + 5 | x | +3 ) − 2   π π π π 2 2 2 −1 −1    ⇒ cot cot 2 2 2 − cos cos = − + − + = − +   2 x 2 + 5| x | +3 2 x 2 + 5| x | +3 9 | x | 2   9| x+  2       0<↓<2   ⇒

x

Que. 63. A.

2

⇒

− 4 | x | +3 = 0 3, −

Solution are

1 3

⇒

= 1,3

x

(

x = ± 1, ±3.

)

, 2 − 3, − 2 + 3 ⇒ Product

= 1.

Que. 64. A.

tan 37.5o

o

1 − cos 75 = =  sin 75o  2

= tan 

75 

o

1−

3 −1 2 2 3 +1

=

2 2 − 3 +1 3 +1

=

(2

)(

2 − 3 +1

)=2

3 −1

2

6 − 4− 2 2

2 2 2

(

6 −3+ 3

) − (2

)

2 − 3 +1

2

∴

a

2 6 −4+2 3−2 2 2

3cos x

+ (3)

4sin x

ad

⇒

= 6; b = 4, c = 3, d = 2

y = (3)

Que. 65. C.

=

bc

2 35 / 2

=

2 3.3. 3

2

=

243

=

12 12

now using AM ≥ GM

⇒ 33cos x + 34sin x ≥ 2 33cos x + 4sin x ≥ 2 3− 5

=

=

4 243

= 6− 4+ 3− 2

= 1.

33cos x

+ 34sin x 2

1/ 2

≥ ( 33cos x.34sin x )

−5 ≤ 3cos x + 4sin x ≤ 5

but

∴ 33cos x + 34sin x ≥ 2 3−5

⇒ a + b = 247.

Domain of f is [ −1, 1] ; f ( x ) = sin x + cos x + tan x + sin −1 x + cos−1 x + tan −1 x

Que. 66. A

f ' ( x ) = cos x − sin x + sec 2 x + 0 + 

>1

1 1+ x2



[1 / 2 , 1]

Hence f ' ( x ) > 0

∴

2

π = + cos1 2

⇒

sin

π 2

π π + π − = + cos1− sin1− tan1 4

4

( A )]

S = cosθ + cos 2θ + ....... + cos nθ

Que. 67. A.

S=

f is increasing ⇒ range is  f ( −1) , f (1) 

f ( x ) |min = f ( −1) = − sin1 + cos1 − tan1 −

M+m

⇒

⇒

where θ = π / 6 and n = 2009

nθ

2 cos ( n + 1) θ θ 2 sin 2

Hence cos ( n + 1) THE “BOND”

θ 2

=0

⇒ now ( n + 1)

θ  2010  π π =  = ( 335 )

⇒

Ans.


S=0

2

 2 6

2



JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 24 of A - 34 Que. 68. A.

= ( 2 + 1)

⇒ 2  2009 C1 ( 2 ) 

2009

2008

− ( 2 − 1)

2009

+ 2009 C3 ( 2 )

2006

+2009 C5 ( 2 )

⇒

= which is an even integer

2004

+ ........ +2009 C2009 ( 2)   0

(A) Comprehesion Type

# 1 Paragraph for Q. 1 to Q. 3 1.

C.

2.

B.

(i )

tan

A

=

2

3.

∆ ∆ r  A s−a = r =  ∴ cot =  s (s − a ) s − a  s 2 r A

s−a+ s− b+s −c

A

∑ cot 2 = ∏ cot 2 =

in any triangle,

A

B

∑ cot 2 cot 2 =

(ii)

A.

cos

A 2

B 2

C

sin

2

sin

A

sin 

Now consider

cos

r

2

B

+

2

2

s2

r

∆

= =

B

C

A

2 B

2 C

2

+ cos cos sin A

s

sin

2

sin

C

A

2

2

+  = 1 ⇒ sin 

2

4s 2

=

=

4∆

abc

C

A

2

2

+ cos cos

= ∏ sin

(a + b + c)

sin

2

 abc  .  4R 

.R  ∆

B 2

........(1)

A 2

B C A B C +  cos + cos  +  sin 2

2

2

2

2

B C B C C A A B C  A B C  A  sin cos cos + sin cos cos cos + cos cos cos  −  sin sin sin  = 1 2 2 2 2 2 2 2 2 2 2  2 2 2 

∴ cos

∴

A 2

cos

B

A

2

C

sin

2

B

∑ cot 2 cot 2 =

(iii)

We have

are not cot f (x) = x

3

A 2 s

B

C

A

2

2

2

+ cos cos sin

A

B 2

A

2

2

sin

B 2

= 1 + ∏ sin

A

∏ sin 2 A ∏ sin 2

1+

  using r = 4R 

and cot

A

C 2

∏ sin 2  = A

1+

r 4R r

=

A 2

4R + r r

.

4R s

∑ cot 2 = ∏ cot 2 = r

, cot

C

+ cos cos

and

A

B

∑ cot 2 cot 2 =

4R + r r

hence an equation whose roots

A A B   cot  x 2 +  ∑ cot cot  x − ∑ 2 2 2  

3 is x − 

A

∏ cot 2 = 0

s  4R + r  π  x − = 0 as A or B C ∈   . r  r  2

− x2 +  r





s 4R + r s − =0 ∴ x = 1 must a root  as cot 2 or cot 2 or cot 2 = 1 ∴ f (1) = 0 ⇒ 1 − + R r r  

⇒

r − 2s + 4R + r = 0

THE “BOND”

A

⇒


B

C

2R + r = s.



JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 25 of A - 34 # 2 Paragraph for Q. 4 to Q. 6 4.

A.

5.

B

6.

C. 3

∏ cosA = 8R

( AH )( BH )( CH ) = 3 i.e. ( 2R cos A )( 2R cos B )( 2R cos C ) = 3 ⇒

3

........(1)

( HD )( HE )( HF) = ( 2R cos B cos C)( 2R cos C cos A)( 2R cos A cos B) = 8R 3 ( cos2 A cos2 B cos2 C ) ....(2) From (1) and (2)

∏ cosA =

∑

cos 2 A

3 8R

. 3

9

∏ ( HD ) = 8R . 64R 3

4R 2 7

=

3

9

=

8R

2

2

2

also ( AH ) + ( BH ) + ( CH ) = 7

3

∴ (1) ÷ ( 3 )

2 2 2 Now we know that, in a triangle ABC cos A + cos B + cos C

14R

= 1 − 2 cos A cos B cos C

6

⇒

7 4R 2

= 1 − 2.

3 8R 3

⇒

7 4R 2

∴ 4R 3 − 7R − 3 = 0 ⇒ ( R + 1) ( 2R + 1)( 2R − 3) = 0

= 1−

3 4R 3 3

∴ R= . 2

Assertion & Reason Type tan B + tan C

Que. 1. (D) A = π − (B + C) tan A = − tan(B + C) =

tan B tan C − 1

⇒ S − 2 is True

hence if A is acute then tan B tan C > 1 if A is obtuse then thanB tan C < 1

⇒ S − 1 is False ⇒ answer is (D) Que. 2. (C) Que. 3. (C)

f (x) = ax 2 + bx + c given f (0) + f (1) = 2 ⇒ f (x) > 0∀x ∈ R ⇒ S − 1 is true. Let f (x) = x 2 − x + 1 ⇒

Que. 4. (A)

⇒

a+b=0

S − 2 is False

cos2 A + cos 2 B + cos 2 C = 1 − 2 cos A cos B cos C

∴

∑ cos

2

A |min = 1 − 2.

1

1

3

4

4

= 1− = .

8

Que. 5. B. Que.6. (a) tan θ − tan α − tan β =

tan α + tan β 1 − tan α tan β

− ( tan α + tan β ) =

tan α + tan β 1 − tan α tan β

( tan α .tanβ ) ⇒ tan θ. tan α .tan β

Que. 7. C. Que. 8. B. Que. 9. D.

Givne tan 2x = 1 ∴ 2x = nπ +

Hence given equation has no solution

π 4

(note that tan 4x is not defined)

∴

Statement - 1 is false and Statement - 2 is true.

Que. 10. A. Que. 11. C.

If it is acute triangle then statement -1 is abviously true

Let A be obtuse say A = 150o

THE “BOND”

∴


B + C = 30 o both angles

< 30o and if C = 30o



JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 26 of A - 34 Now

cot A and cot ( B + C ) will be of equal magnitke but opposite sign, As cot θ is decreasing

hence, cos B + cos A alone is +ve

Que. 12. B

∴ cos A + cot B + cot C > 0 . (∴ chord BD subtends equal angle in same segment)

∠BPD = ∠BCD = 60°

∴

∠APB + ∠DPB = 180° A, P, D are collinear

|||ly

B, P, E and C, P, F are also collinear

hence AD, BE, CF are concurrent at P.

Que. 13. A.

∴

∆BHN and BDN are congruent ⇒

HN = ND = 2R cos Bcos C

HD = 4R cos Bcos C

More than One May Correct Type Que. 1. (A,B,C,D) 2 cos θ + 2 2

∴

= 3sec θ

2 cos 2 θ + 2 2 cos θ − 3 = 0 cos θ =

−2 2 ± 32 4 1

∴

cos θ =

∴

θ=

⇒

sin θ = −

THE “BOND”

π 4

or

2

−

or 1 2

=

−2 2 ± 4 2 4 cos θ = −

3 2

( rejected )

π 4

; cot θ = − 1;


tan θ = 1

and

cos θ = −

1


2


JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 27 of A - 34 Que. 2 (B,C) (A) =

1 2

sin150

o

=

1 4

⇒

rational

⇒

irrational

(B)

2 + log 2 7

(C)

log3 6 = 1 + log 3 2

(D)

8−1/ 3

= 2−1 =

1 2

⇒

⇒

irrational

irrational

Que. 3. (B,C,D) Opposite angles of a cyclic quadrilateral are supplementary

 11π   5π   π   π  1 π 1  sin   = sin   cos   = sin   = ∈ Q  12   12   12   12  2  6  4

sin 

Que. 4. (A,B,C,D) (A)

1 16  9π   4 π   π   π =− = −4 ∈ Q  sec   = − cos ec   sec   = o o  19   5   10   5  sin18 cos36 ( 5 − 1)( 5 + 1)

(B).

cos ec 

(C).

sin 4 

(D).

2cos 2

1 1 3 π 4 π 2 π 2 π 2 π  + cos   = 1 − 2sin   cos   = 1 − sin   = 1 − = ∈ Q 2 4 4 8 8 8  8  4

Que. 5. (A,C)

1 2

ra +

π 9 1 2

.2 cos 2

rb =

1 2

2π 9

.2cos2

4π 9

1

= 8 ( cos 20 o.cos 40 o.cos80 o ) = ∈ Q 8

ab sin C ⇒ r(a + b) = 2∆ ⇒ r =

2∆ a+b

......(1)

C

A

B R

R

A

∴ r=

2abc 4R ( 2R sin A + 2R sin B )

form (1)

r=

1 2. ab sin C 2 a+b

=

B

I

=

abc 4R

2

2ab sin

( sin A + sin B ) C

cos

2 a+b

C 2

=

⇒ ( A ) also x =

2ab cos

2ab a+b

cos

C 2

C

2 .sin C a+b 2

= x sin

C 2

Que. 6. (A,C) sin x − 3sin 2x + sin 3x = cos x − 3cos 2x + cos 3x or 2sin 2x cos x − 3sin 2x = 2 cos x − 3cos 2x sin 2x ( 2 cos x ) − 3 = cos 2x [ 2 cos x − 3] ⇒ ( sin 2x − cos 2x)( 2 cos x − 3) = 0 but 2cos x − 3 ≠ 0

as cos x ≤ 1 hence, sin 2x − cos 2x = 0 ⇒ tan 2x = 1 ⇒ 2x = nπ + π / 4 or x =

THE “BOND”



nπ 2

π + ⇒ a, b, c 8



(sin

Que. 7. (A,B,D)

2

3

x+ cos 2 x ) - 3sin 2 xcos 2 x (sin 2 + cos 2 x ) > 5

5

8

8

5 8

= 1 − 3sin 2 x cos 2 x > ⇒ 1 − > 3sin 2 x cos 2 x ⇒

π π π π > 3sin 2 x cos 2 x ⇒ 1 − 2sin 2 2x > 0 ⇒ cos 4x > 0 ⇒ 4x ∈  − ,  ⇒ 4x ∈  2πr − , 2π r +  8 2 2 2 2 3



 nπ π nπ π  − , + n∈I  2 8 2 8

x ∈







now verify .

Que. 8. (A,B,D) sin 24o cos 60 o − cos 24o sin 6 o

= −1 (A) 1 (B) 3 (C) sin 21o cos 39o − sin 39 o cos 21o = sin ( −18o )

Que. 9. (B,C,D)

cos 2 α − sin α

Que. 10. (A,C)

(A)

Que. 11. B,D.

tan ( α + β ) =

Que. 12. A,B,C.

(A).

(B).

sin 2θ =

(C).

tan3θ =

(D).

sin θ =

Que. 13. A,B,C,D.

sin (18o )

sin α cos α

1 + tan

3

8

sin θ cos3θ

2

θ

and tan γ =

sin2θ

; cos 2θ =

⇒

–1

= 2cot2α

1 − cos 2θ

tan α =

2 tan θ

1

15

(D)

8 15

∴ α+β+ γ =

π 2

⇒ (B) and (D)

⇒ (A) is correct.

1 − tan 2 θ 1 + tan

2

θ

; tan 2θ =

2 tan θ 1 − tan 2 θ

⇒ (B) is correct.

(C) is correct.

which is rational but cos 3θ = cos θ ( 4 cos 2 θ − 3) which is irrational ⇒ (D) is correct.

3θ = nπ + ( −1)

n

∴ 3θ = 3α;

( 3α )

3θ = π − 3α;

3θ = − π − 3α or 3θ = 2π + 3α;

3θ = − 2π + 3α

Hence θ = α; θ =

π

2π π   2π  − α; θ = −  + α  ; θ =  + α  ; θ = − + 3α 3 3 3   3 

π   2π  ⇒ cos θ = cos  ± α  or cos θ = cos  ± α  ⇒ (A), (B), (C) and (D) all are correct. 3   3  Que. 14. A,B. Given quadratic equaton is an identity ∴ cosec2 θ = 4 and cot θ = − 3 ⇒ cosec θ = 2 or

− 2 and tan θ = −

Que. 15. A,B,C.

1 3

⇒

5π 6

or

11π 6

Making quadrtic in sine from a cos θ + b sin θ + c, we get

( a 2 + b2 ) sin 2 θ − 2bc sin + c2 THE “BOND”

θ=

− a2 = 0


α ................ (1) β ||Chem. by Pavan Gubrele||


JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 29 of A - 34 2bc

⇒ sin α + sin β =

a

+ b2

2

⇒ (A) is correct ⇒ sin α + sin β =

c2 − a 2 a 2 + b2

⇒ (B) is correct

Making quadratic equation in cos, we get (changing a and b)

α β

( a + b ) cos θ − 2ac cos θ + c − b = 2

2

2

2

⇒ cos α + cos β =

2bc a 2 + b2

⇒ (C) is correct ⇒ cos α + cos β =

∑ cos 3A = 1 ⇒ sin

Que. 16. B,D.

A

alos r = ( s − a ) tan

2

3 −1 2 2 sin x

∴ 2x = x + Que. 18. A.

π 12

+

3A

or ( s − b ) tan

r = 3 ( s − a ) or

Que. 17. A,D.

2

2

.sin

B 2

2 2 cos x

π 12

3C 2

=0 ∴ A=

C

3 ( s − c)

⇒

= 2 ⇒ sin

or 2x = π − x −

2

.sin

or ( s − c ) tan

3 ( s − b ) or

3 +1

3B

π 12

π 12

a 2 + b2 2π 3

⇒ (D) is correct

or B =

2π

or C =

3

2π 3

⇒ (B)

2

cos x + cos

⇒ x=

c2 − b 2

π 12

(D)

 

sin x = sin 2x ⇒ sin 2x = sin  x +

or 3x =

11π 12

∴ x=

π 12

or

11π 36

π 

12 

⇒ A, C.

Square and adding 9 + 16 + 24 sin ( A + B ) = 37 ⇒ 24 sin ( A + B ) = 12 ⇒ sin ( A + B) = 1

2

1

⇒ sin C = ; C = 30o Or 150o if C = 150o then even of B = 0 and

A = 30 o the quantity

2

1 1 3sin A + 4 cos B ⇒ 3. + 4 = 5 < 6 hence C = 150o is not possible 2 2

⇒ ∠C = 30o only

cos 3θ = cos 3α put n = 0,1 ⇒ 3θ = 2nπ ± 3α

Que. 19. A,B,C,D.

∴ 3θ = 3α or − 3α or 2π + 3α or 2π − 3α θ = α or − α or

2π 3

+ α or

2π 3

−α ⇒

if n = −1 ⇒ 3θ = −2π ± 3α ⇒ θ = −

2π 3

(A), (C), (D) are correct.

±α

π π π 2π    2π ± α  = − sin  ± α  = − sin  π − ± α  = − sin  π −  ± α   = − sin  ± α  3  3   3     3   3

sin θ = sin  −

hence (B) is not correct. Que. 20. B, C, D i.e.

use power of point B w.r.t. the circle passing through AC1GB1

BC1 × BA = BG × BB1 ⇒

THE “BOND”


c 2

2

c2

2

2

× ( mb ) ⇒

× c = BB1 × BB1 ⇒ 3 3 3


c2

2  2c 2 + 2a 2 − b 2 

=  3 3

4

 



⇒

2a 2

⇒

= b2 + c2 Ans.

B, C, D are the answers.

Match Matrix Type Que. 1. A - R. A.

B - S.

C- P.

D - Q.

cos2 2x − sin 2 x = 0 ⇒ cos 3x.cos x = 0 ⇒ cos 3x

x = ( 2n −1) (2n − 1)

π 6

π 6

or x = ( 2n −1)

π 2

∴

x−

3 tan 2 x −

C.

hence general solution is (2n − 1)

π 6

as ( 2n − 1)

π 2

π 2

is contained in

⇒ (R)

cos x + 3 sin x = 3 ⇒

B.

= 0 or cos x = 0 ⇒ 3x = (2n − 1)

∴ or

(

π 3

= 2nπ +

π 6

cos x 2

+

3 2

or 2nπ −

sin x =

π 6

π π π π  ⇒ cos  x −  = cos ⇒ x − = 2nπ ± 2 3 6 3 6  3

⇒ x = 2nπ +

π 2

or x = 2nπ +

π 6

⇒ (S)

)

3 + 1 tan x + 1 = 0 ⇒ 3 tan x ( tan x − 1) − ( tan x − 1) = 0 ⇒ ( tan x − 1)

⇒

tan x − 1 tan x =

x = nπ +

(

)=0

3 tan x − 1

π 4

 ⇒ (P)

1

⇒

3

x = nπ +

π 6 

D.

tan 3x − tan 2x − tan x = 0 or tan x.tan 2x.tan 3x = 0 ⇒ x = nπ or

⇒

general solution

nπ 3

are nπ is contained in

nπ 3

⇒

nπ 2

(rejected) or

nπ 3

(Q)

Subjective Type ( Up to 4 digit) Que. 1.

π 17

= θ; 17θ = π ⇒ 2 cos θ.cos 9θ + cos 7θ + cos 9θ ⇒

cos(100)θ + cos8θ + cos 7θ + cos9θ = 0

Sum of cosines of supplymentary angles is zero. Que. 2

tan α

2sin α cos β − 2sin α 2cos α cos β − 2cos α

THE “BOND”

=

2sin α (cos β − 1) 2cos α (cos β − 1)


= tan α Hence proved. ||Chem. by Pavan Gubrele||


JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 31 of A - 34 Que. 3. Multiple numerator and denominator by 2 cos θ − cos 3θ + cos 3θ − cos 9θ + cos 9θ − cos17θ

∴

sin 3θ − sin θ + sin 9θ − sin 3θ + sin17θ − sin 9θ

=

cos θ − cos17θ sin17 θ − − sin θ

=

2sin 9θ sin 4θ 2sin 4θ cos9θ

= tan 9θ = tan kθ ⇒

k = 9.

3 2 7 3 ; 8 16

Que. 4.

8

8

x  x x x  x x x x    y =  cos  −  sin  =  cos 4 − sin 4  cos 4 + sin 4  = cos x 1 − 2 sin 2 cos 2  2  2 2 2  2 2 2 2    1 = cos x 1 − sin 2 x   2  3 3 2 π 1  1 1 1 3 . = . = = 1 − .  = 2 2 8 2 2 4 4 2 4

(i)

y

(ii)

y

3 1 1 7 3 π = 1 − ;  =  6  2  2 4  16 sin A

Que. 5. (229) cos B

a

4 2 + 6 2 − 52 2.5.6

deividing by 16

=

= 9

16

cos A 12

sin B b

=

sin C c

; cos C =

=

cos B 9

4 2 + 52 − 6 2

=

5

⇒ a = 4k, b = 5k, c = 6k ∴ cos A =

2.4.5 cos C

1

cos A

8

3/ 4

= hence

=

cos B 9 /16

=

2

+ 62 − 4 2 2.5.6

3

= ; 4

cos C 1/ 8

∴ x = 12, y = 9 and z = 2.

2

Que. 6. A = 3; B = 1; C = −2; D = 2 ⇒ 3 + 1 − (−2) ÷ 2 = 5. Que.

E = ( cos2 x + sin 2 x )

7.

2

− 2sin2 x cos2 x − k2 ( cos2 x − sin2 x)

2

= 1− 2 − sin 2 x cos 2 − k 2 ( cos 2 x + sin 2 x ) − 4sin 2 x cos 2 x  = (1 − k 2 ) − 2 sin 2 x cos 2 x (1 − 2k 2 ) for this to be independent of x, 1 − 2k 2 = 0 ⇒ k = is

1 2

Que. 8. cot

1 2

or −

1 2

Note :The value of expression for this value of k

.

π 24

= cos 7.5o =

1 + cos15 o

sin15

6+ 2

1+

o

=

4 6− 2

=

4+ 6 + 2 6− 2

=

(4 +

6+

2

)(

6+ 2

)

4

4

=

4

(

Que. 9. L =

6+ 2

) + ( 8 + 4 3) = 4

6 + 2 +2+ 3

= 2 + 3 + 4 + 6 ∴ p + q + r + s = 15.

−2 cos 4θ sin 2θ + 2sin 4θ sin 2θ 2sin 2θ ( sin 4θ − cos 4θ ) = = 2sin 2θ sin 4θ − cos 4θ sin 4θ − cos 4θ

THE “BOND”




JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 32 of A - 34 5 +1

o o o If θ = 27 , L = 2sin 54 == 2 cos 36 ⇒ L =

M=

tan x tan 2x tan 2x − tan x

=

2tanx tan x. 1 − tan 2 x 2tanx − tan x 1 − tan 2 x

o o when x = 9 , M = sin18 ⇒ M =

N=

1 − cos 4α sec 2α − 1 2

+

1 + cos 4α cos ec 2α − 1 2

=

=

5 −1 4

2

2 tan x 2 − (1 − tan x ) 2

=

2 tan x 1 + tan 2 x

= sin 2x

.

2 sin 2 2α.cos2 2α

(1 − cos 2α ) 2

+

2 cos2 2α.sin 2 2α

(1 − sin 2α) 2

= 2 ( cos 2 2 α + sin 2 2 α) ⇒ N = 2.

 5 + 1  5 − 1    4  (2) = 1. 2   

∴ LMN = 

Que. 10. cos ( α + β) =

4 5

⇒ tan ( α + β) =

3 4

⇒ sin ( α − β) =

5 13

⇒ tan ( α − β) =

5 12

now 2β = ( α + β) − ( α − β)

3 5   4 − 12  16 tan ( α + β ) − tan ( α − β ) = . = tan 2β = 1 + tan ( α + β ) . tan ( α − β )

3 5 1+ . 4 12

63

2

2

Que. 11. 2 (1 − sin 2 x ) − sin 4 x + k = 0 put sin 2 x = t, t ∈ [ 0,1] ⇒ 2 (1 − t ) − t 2 + k = 0 ⇒ t 2 − 4t + k + 2 = 0 since sum of the roots is 4 ⇒ one root in (0,1) and other greater then 1 as shown y

x

O now

f (0) ≥ 0 and f (1) ≤ 0 ⇒ k + 2 ≤ 0 and k − 1 ≥ 0 ⇒ k ∈ [ −2,1].

Alternatively : 2 cos4 x − sin 4 x + k = 0 ⇒ cos4 x + ( cos4 x − sin4 x ) + k = 0 ⇒ cos4 x + cos 2x + k = 0 2

 1 + cos 2x  2 2   + cos 2x + k = 0 ⇒ 1 + cos 2x + 6 cos 2x + 4k = 0 .....(A) ⇒ cos 2x = t ⇒ t + 6t + 1 + 4k = 0 2   2

2

2

⇒ ( t + 3) = 8 − 4k ⇒ ( t + 3) max. = 16 ⇒ ( t + 3)min. = 4 ∴ 4 ≤ 8 − 4k ≤ 16 ⇒ −4 ≤ −4k ≤ 18 ⇒ 1 ≥ k ≥ −2 Alternatively : After step (A)

cos2x

=

−6 ± 36 − 16k − 4 2

=

−6 ± 32 − 16k 2

think !) cos 2x = −3 + 2 2 − k or − 3 − 2 2 − k (rejected,

⇒ −1 ≤ −3 + 2 2 − k ≤ 1 ⇒ 2 ≤ 2 2 − k ≤ 4 ⇒ 1 ≤ 2 − k ≤ 2 ⇒ 1 ≤ (2 − k) ≤ 4 ⇒ − 1 ≤ − k ≤ 2 ⇒ 1 ≥ k ≥ − 2 2π  4π  2π  4π    ; c = sin  θ + + sin  θ +  Hence, a + b + c = sin θ + sin  θ + Que. 12. Let a = sin θ; b = sin  θ +    3  3  3  3     

THE “BOND”





    2π  2π   π  π  π  = sin θ + sin  − θ  − sin  + θ  use : sin  π +  = sin  π − θ +   = sin  − θ  (using C-D) 3 3 3 3 3 























π = sin θ − 2 cos sin θ = sin θ − sin θ = 0 since a + b + c = 0 hence a 3 + b3 + c3 = abc 3 ∴ sin 3 θ + sin 3  θ +

2π 

4π  π  π   2π  3 2  + sin  θ +  = −3sin θ sin  − θ  sin  + θ  = − 3sin θ  sin − sin θ  3  3  3  3  3   



=−

3

3 3sin θ − 4 sin θ ) = − sin 3θ. ( 4 4 3

Que. 13. Tr =

sin 2r −1 r −1

cos 2 .cos 2

=

r

sin ( 2r

H.P.

− 2 r−1 )

r −1

cos 2 .cos 2

=

r

sin 2r cos r −1 − cos 2 r sin 2 r−1 r −1

cos 2 .cos 2

r

= tan 2r − tan 2r −1

n

∴ Sum = ∑ ( tan 2r − tan 2r −1 ) = tan 2 − tan1+ tan 22 − tan 2 + tan 23 − tan 22 + .........+ tan 2n − tan 2n−1 r =1

Sum = tan 2n

− tan (1)

Que. 14. (0250.00) Dividing

by

15 tan 4 α + 10 = 6sec 4 α ⇒ 15 tan 4 α + 10 = 6(1 + tan 2 α ) 2

cos4 α 2

⇒ 9 tan 4 α − 12 tan 2 α + 4 = 0 ⇒ ( 3 tan 2 α − 2 ) = 0 ⇒ tan 2 α =

2 3 3

 3  2 8cosec α + 27 sec α ⇒ 8 (1 + cot α ) + 27 (1 + tan α ) ⇒ 8 1 +  + 27  1 +  ⇒ 125 + 125 = 250.  2  3

Now

6

6

3

2

2

3

x + sin y = 2008

Que. 15. (2008).

∴ y=

π 2

Subtract

x + 2008cos y = 2007 sin y − 2008cos y = 1

and x = 2007 ⇒ x + y = 2007 +

π 2

⇒ sin y = 1+ 2008cos y This is possible only if cos y = 0

⇒ [ x + y ] = 2008.

2 sin x sin1 = cos ( x − 1) − cos ( x + 1)

Que. 16. (92) 44

∴ S∑ cos ( x − = 1) − cos ( x + 1)  1 + sec ( x − 1) .sec ( x + 1)  x =2

  1 1   = ∑ cos ( x − 1) + − cos(x − 1) −  cos(x + 1) cos ( x − 1)  x =2  

 1 − cos2 ( x + 1) 1 − cos 2 (x − 1)  = ∑  −  cos ( x + 1) cos(x − 1)  x=2 

44

44

2 2 2 2 2 2 2 2  sin 2 (x + 1) sin 2 ( x − 1)  sin 3 sin 1 sin 4 sin 2 sin 4 sin 3 sin 44 sin 42 ........ + = ∑  − − + − + − − ∴ S = cos ( x − 1)  cos 3 cos1 cos 4 cos 2 cos 5 cos 3 cos 44 cos 42 x = 2  cos ( x + 1) 44

+

sin 2 44 cos 44

−

sin 2 42 cos 42

+

sin 2 45 cos 45

−

sin 2 43 cos 43

⇒ S=

sin 2 44 cos 44

+

sin 2 45 cos 45

−

sin 2 1 sin 2 2 cos1

−

cos 2

φ2 ( θ n ) S=− + − + which resembles 4 term of ∑ ( −1) ψ ( θn ) n =1 cos1 cos 44 cos 2 cos 45 sin 2 1

∴

sin 2 44

sin 2 2

4

sin 2 45

n

θ1 + θ 2 + θ 3 + θ 4 = 1+ 2 + 44 + 45 = 92.

THE “BOND”




JEE/AIEEE MATHS by SUHAAG SIR Bhopal, Ph. (0755)32 00 000 Teko Classes IIT www.tekoclasses.com Question. & Solution. Trigo. Page: A - 34 of A - 34 f (x) = cos −1 ( sin x ) =

Que. 17. (3388) g(x) = sin

−1

π 2

− sin −1 ( sin x ) ...........(1)

π ( cos x ) = − cos −1 (cos x)

and

............... (2) both f(x) and g(x) are periodic with period

2

−1 −1 2π. The graphs of sin ( sin x ) and cos ( cos x ) as follows

π  2 − x if x ∈ [ 0, π / 2]  π π 3π
π  2 − x if x ∈ [ 0, π ] g(x)   x − 3π if x ∈ [ π, 2π ]  2

Now

Area enclosed between the two curves is the area of the rectangle ABCD in one pepriod. AD =

and

DC = 2π

DC =

2π

Now

and

π2

now

THE “BOND”

4

+

π2 4

=

π2 2

∴ A = 7.π2 = 7 π2


=

π 



2⇒

 

π 2

. 2 [= π

( in [ −7π, 7π])

2

49A = 49.7π


2

= 7.22.22 = 7.484 = 3388.


Trigonometry

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