Inverse Trigo

INVERSE TRIGONOMETIC

23 sec x =   x = sec  –1

where, –  < x  –1 or 1  x <  and 0   <

(vi)

  or < 2 2

Definition of cosec–1 x :    cosec–1 : (–, –1]  [1, )   ,  – {0}  2 2

such that cosec–1 x =   x = cosec  where

– < x  –1 or 1  x <  and –

    < 0 or 0 <  2 2

Summary Function

Domain of the function

Principal value branch

  y 2 2

1.

y = sin–1x

–1  x  1

–

2.

y = cos–1 x

–1  x  1

0y

3.

y = tan–1 x

– < x < 

–

4.

y = cot–1 x

– < x < 

0
5.

y = sec–1 x

x  –1 or x  1

0  y  , y 

6.

y = cosec–1 x

x  –1 or x  1

–

 
 2

  y ,y0 2 2

Note : Unless otherwise stated sin–1 x, cos–1 x, tan–1 x, cot–1, x sec–1 x and cosec–1 x will mean their principal branches. 2.4

GRAPHS OF PRINCIPAL BRANCHES OF INVERSE TRIGONOMETRIC FUNCTIONS

Y y=1

(i)

Graph of y = sin x Domain = R = (– , ) and Range = [–1, 1]

 2

– 3  2



 2

 2

O

y = –1

Graph of y = sin–1 x Domain = [–1, 1]    Range =  ,   2 2

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

3 2

2

X

INVERSE TRIGONOMETIC

25 (iv)

Graph of y = cot x Y

Domain = R – {x : x = n , n  Z} Range = (–, ) –



 2



O  2

X

Y 

Graph of y = cot–1 x

3 2

Domain = (–, ) Range = (0, )

 2

 4 O

X

Y (v)

Graph of y = sec x    Domain = R – x : x  (2n  1) , n  Z  2     O  ) Range = (–, -1]  [1, 2 2



X

3 2

Y

Graph of y = sec–1 x



Domain = (–, -1]  [1, )

     Range = 0,    ,  = [0, ] –   2 2 2     2

–1

(vi)

O

1

X

Y

Graph of y = cosec x Domain = R – {x : x = n , n  Z} Range = (–, –1]  [1, )

 –

Graph of y = cosec–1 x Domain = (–, -1]  [1, )

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 2 O

 2



X

INVERSE TRIGONOMETIC

27 Property III : (i)

sin–1 x = cosec–1 cosec–1 x = sin–1

1 , –1  x  1 and x  0 x 1 , x  –1 or x  1 x

1 , –1  x  1 x 1 sec–1 x = cos–1 , x  –1 or x  1 x

(ii)

cos–1x = sec–1

(iii)

 1  1 tan–1x = cot–1   , x > 0 = –  + cot–1   , x < 0 x x

(iv)

 1  1 cot–1 x = tan–1   , x > 0 =  + tan–1   , x < 0 x x

Property IV : (i) sin–1 (– x) = – sin–1 (x), (ii) cos–1 (– x) =  – cos–1 x, (iii) tan–1 (– x) = – tan–1 x, (iv) cot–1 (– x) =  – cot–1 x, (v) sec–1 (– x) =  – sec–1 x, (vi) cosec–1 (– x) = – cosec–1 x,

for all x  [–1, 1] for all x  [–1, 1] for all x  R for all x  R for all x  (–, –1]  [1, ) i.e., for all |x|  1 for all x  (–, –1]  [1, ] i.e., for all |x|  1

Property V :

 , 2

(i)

sin–1 x + cos–1 x =

(ii)

tan–1x + cot–1 x =

(iii)

sec–1 x + cosec–1 x =

 , 2

for all x  [–1, 1] for all x  R

 for all x  (–, –1]  [1, ) i.e., for all |x|  1 2

Property VI :

(i)

  xy   , tan 1   1  xy    1 x  y   , tan–1 x + tan–1 y =    tan   1  xy    1 x  y   ,     tan    1  xy 

if xy  1 if x  0, y  0 and xy  1 if x  0, y  0 and xy  1

Property VII :

xy tan–1 x – tan–1 y = tan–1 1  xy , if xy > –1 This result can be established by putting –y in place of y in the results of property V using the fact that tan –1 (– y) = – tan–1 y.   xy   , tan 1   1  xy    1 x  y   , tan–1 x – tan–1 y =    tan   1  xy    1 x  y   ,    tan    1  xy 

if xy  1 if x  0, y  0 and xy  1 if x  0, y  0 and xy  1

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INVERSE TRIGONOMETIC

29

(ii)

  1 x2   , if 0  x    cos 1   1 x2     2 tan–1 x =  2  1  x  1   , if    x  0  cos  2 1  x   

x sin–1 x = cos–1

Property XIV : (i)

–1 1 x 2 = tan

= cot–1

1 x2

 1  1 x2 = sec–1  2 x  1 x

  1  –1  = cosec  x  

Where x  0

(ii)

–1

cos x = sin

–1

1 x

2

= tan

 x  = cot–1  2  1 x

–1

 2  1 x  x 

   

 1  –1 –1  = sec x = cosec 

 1   2  1 x

   

 2  1 x  x 

   

Where x > 0

(iii)

 x  tan–1 x = sin–1  2  1 x

= cot

–1

  –1  = cos 

 1   = sec–1 x

 1   2  1 x

1 x

2

   

= cosec

–1

Property XV :

(i)

 1 1 2 if  x sin (2x 1  x ), 2  1  1 2  x 1 2 sin–1 x =    sin (2x 1  x ), if 2      sin (2x 1  x 2 ), if  1  x    

1 2

1 2

Property XVI : (i)

 cos 1(2x 2  1) , if 0  x  1 2 cos–1 x =  2  cos 1(2x 2  1) , if  1  x  0

Property XVII :

 sin 1(3 x  4 x 3 ) ,   1 3 3 sin–1 x =    sin (3 x  4 x ) ,      sin 1(3 x  4 x 3 ) , 

if  if

1 1 x 2 2

1  x 1 2

if  1  x  

1 2

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INVERSE TRIGONOMETIC

31

SOLVED PROBLESM Ex.1

Write each of the folloiwng functions in the simplest form :

 x  (i) tan  2 2  a x –1

Sol.

(i)

  , | x | < a 

a 3

Let x = a sin . Then,

 x  tan–1  2 2  a x

Thus, tan (ii)

 3a2 x  x3  a  (ii) tan  3 x 2  , a > 0; a  3 ax 3   –1

–1

  –1  = tan 

 x   2 2  a x

  a sin   a sin   x   –1  –1 –1     2 2 2  = tan  a cos   = tan (tan ) =  = sin  a   a  a sin  

 x  –1  = sin  a  

Let x = a tan . Then

tan

–1

 3a 2 x  x 3    –1  a3  3ax 2  = tan  

 3a3 tan   a3 tan3     –1  a3  3a3 tan 2   = tan  

 3 tan   tan 3      1  3 tan 2    

x = tan–1 (tan 3) = 3 = 3 tan–1   a

 3a3 x  x 3   –1 Thus, tan–1  3 2  = 3 tan a  3 ax  

Ex.2

Sol.

x   a

Simplify : (i)

 a cosx  b sinx  a  , if tan–1  tan x > –1 b cos x  a sin x b  

(ii)

1 tan 2

(i)

We have

tan

–1

2  2x 1 1  y  sin 1  cos  1 x2 1  y2 

  ,| x | < 1, y > 0 and xy < 1  

 a cos x  b sin x   a cos x  b sin x  b cos x –1   = tan   b cos x  a sin x   b cos x  a sin x  b cos x 

   = tan–1   

 a    tan x  b    1  a tan x    b  

 pq   , where p = a and q = tan x = tan–1  b  1  pq  a = tan–1 p – tan–1 q = tan–1   – tan–1 (tan x) = tan–1 b

(ii)

a   –x b

Let x = tan  and y = tan . Then

1 tan 2

2  1 2x 1 1  y   sin  cos  1 x2 1  y 2  

 1 1 2x 1 1  y 2  1 1    cos 1 (2 tan 1 x )  [2 tan 1( y )] = tan  2 sin 2 2  = tan  2 2 2 1 x 1 y    

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INVERSE TRIGONOMETIC

33 3     =– Hence, tan–1  tan 4   4

(iii)

We know that cos–1 (cos x) = x 

7   7  = cos–1  cos 6   6

But,

7  [0, ] which is the principal branch of cos–1 x 6

However, cos

7 = cos 6

5   5 5  2   = cos and  (0, ) 6   6 6

7   5  = Hence, cos–1  cos 6   6

Ex.5

Prove that

 1 x  1 x  1  1   tan–1  = – cos–1 x, for – x1  2 2 2  1 x  1 x  Sol.

Let x = cos 2 . Then,

1  cos 2 =

1 x =

1  cos 2 =

2 cos2  =

2 sin2  =

and

1 x =



 1 x  1 x  LHS = tan–1   1 x  1 x

2 cos 

2 sin 

   

 2 cos   2 sin    cos   sin     –1 = tan–1   = tan  cos   sin   2 cos   2 sin   

 1  tan    = tan –1 = tan –1   1  tan  

   tan    4    = Ex.6

Prove that tan–1

Sol.

  1 –= – cos–1 (x) = RHS 4 4 2

x =

 1 x  1  , x  (0, 1) cos–1  2  1 x 

Let x = tan2 . Then, x = tan ; and so tan–1 ( x ) = tan–1 (tan ) = 

Further, RHS =

Ex.7

Sol.

1 cos–1 2

 1 x  1   = cos–1 1  x   2

 1  tan 2   1 1   –1  1  tan 2   = 2 cos (cos 2) = 2 × 2 =   

Hence, LHS = RHS Solve each of the following for x :

8  15    = (i) sin–1   + sin–1  x  x  2

(ii) sin–1 (x) – cos–1 (x) = sin–1 (3x – 2), if x > 0

 8 (iii) tan–1 (x + 1) + tan–1 (x – 1) = tan–1    31

(iv) tan–1 2x + tan–1 3x =

(i)

8 We have sin–1   + sin–1 x



 15     =  x  2

 15  8  sin–1   = – sin–1    x  x 2

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 4

INVERSE TRIGONOMETIC

35

xy holds only when xy < 1 1  xy

tan–1 x + tan–1 y = tan–1

Thus, when x = –1, (2x) (3x) = (–2) (–3) = 6 > 1 So, we reject x = –1 and accept x =

Ex.8

Solve for x : 2 tan–1 x = sin–1

Sol.

Let a = tan  and b = tan . Then,

 2a   = sin–1 sin–1   1  a2 

1 6

1 b2 2a –1 2 – cos 1 a 1 b2

 2 tan      1  tan 2  

= sin–1 (sin 2) = 2 and

cos

–1

 1  b2    –1  1  b 2  = cos  

 1  tan2      1  tan2    

= cos–1 (cos 2) = 2 Hence, RHS = 2 – 2 = 2 ( – ) = 2 (tan–1 a – tan–1 b) So, the given equation implies 2 tan–1 x = 2 (tan–1 a – tan–1 b), –1 < a < 1, –1 < b < 1  tan–1 x = tan–1 a – tan–1 b  ab   ; ab > –1 = tan–1   1  ab 

 Ex.9

ab ; ab > –1 1  ab

Solve : (i) 2 tan–1 (cos x) = tan–1 (2 cosec x) (ii)

Sol.

x=

(iii) (i)

1 x 1 = tan–1 x (x > 0) 1 x 2 sin [2 cos–1 {cot (2 tan–1 x)}] = 0 We have 2 tan–1 (cos x) = tan–1 (2 cosec x) tan–1



 2 cos x   2  –1    tan–1  2  = tan  sin x   1  cos x 

2 cos x

when

2 sin x 1  cos x  sin x (cos x – sin x) = 0 sin x = 0, x = n, n  Z

when

cos x – sin x = 0 or tan x = 1, x = n +

(ii)

We have, tan–1



(iii)

2

=

1 x 1 = tan–1 x 1 x 2

1 tan–1 x 2

sin x cos x = 1 – cos2 x = sin2 x



sin x = 0 or cos x – sin x = 0

 4 (x > 0)



tan–1 1 – tan–1 x =



 1 3 = tan–1 x + tan–1 x = tan–1 x 4 2 2



tan–1 x =

We have,



( tan–1

xy = tan–1 x – tan–1y) 1  xy

1 2    × =  x = tan = 3 3 4 6 6 –1 –1 sin [2 cos {cot (2 tan x)}] = 0

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INVERSE TRIGONOMETIC

37

UNSOLVED PROBLEMS EXERCISE – I Q.1

Evaluate the following :– (i)

Q.2

4    cos–1  cos 3  

 bc    + tan–1  1  bc 

 a3  b3   –1 = tan–1  3 3  + tan 1  a b  

 c a     1  ca 

 b3  c 3    –1  1  b3c 3  + tan  

 c 3  a3     1  c 3a3   

Prove that sin–1

Q.4

(iii)

Prove that  ab   + tan–1 tan–1   1  ab 

Q.3

cos–1 (cos 10)

(ii)

12 4 63 + cos–1 + tan–1 = 13 5 16

Prove that

 ab x   b  a cos x    for 0 < b  a and x  0 2 tan–1  a  b tan 2  = cos–1   a  b cos x   

Q.5

If tan–1x + tan–1y + tan–1z =

 , prove that 2

xy + yz + zx = 1 Q.6

If cos–1x + cos–1y + cos–1z = , prove that x2 + y2 + z2 + 2xyz = 1

Q.7

If sin–1x + sin–1y + sin–1z = , prove that x 1 x 2 + y 1 y 2 + z 1 z 2 = 2xyz

Q.8

Solve : tan–1 2x + tan–1 3x =

Q.9

Solve : sin–1x + sin–1y =

 4

2 3

cos–1x – cos–1y =

 3

Q.10

If sin [2 cos–1 {cot (2 tan–1 x)}] = 0, find x.

Q.11

If –1  x, y, z  1, such that sin–1x + sin–1y + sin–1z =

3 , 2

find the value of x2000 + y2001 + z2002 – Q.12

9 x

2000

y

2001

 z 2002

Prove that tan–1

1  n  1 1  + tan–1 + ........... + tan–1 2 = tan–1  n2 3 7 n  n 1

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1 5  tan   cos 1   3   2  

INVERSE TRIGONOMETIC

39

BOARD PROBLES EXERCISE – II Q1.

 1 Prove that following : tan–1   + tan–1 3

Q2.

2  2     + sin–1  sin  Using principal value, evaluate the following : cos –1  cos 3  3   

Q3.

Q4.

Q5.

 1   + tan–1 5

 1   + tan–1 7

 1    = . 8   4

 1  sin x  1  sin x  x     Prove the following : cos–1  = , x   0,   2  4  1  sin x  1  sin x  OR Solve for x : 2 tan–1(cos x) = tan–1 (2 cosec x) 7    Write the principal value of cos–1  cos 6  

[CBSE 2008]

[CBSE 2009] [CBSE 2009]

 2x   = tan–1 Prove the following : tan–1 x + tan–1   1 x2  OR

 3x  x3     1  3x 2   

1 x2

Prove the following : cos [tan–1 (sin {cot–1 x)}] =

[CBSE 2010]

2  x2

Q6.

4   . Find the value of sin–1  sin 5  

Q7.

 1 x  1 x  1  1  = Prove that : tan–1  – cos–1 x, – x1 1  x  1  x 2 4 2  

Q8.

6  1 3 1 3  Prove the following : cos sin 5  cot 2   5 13   Show that :

Q9.

[CBSE 2008]

[CBSE 2010 ]

3 4  7 1 tan  sin1   4 3 2

[CBSE 2011]

[C.B.S.E. 2012]

OR

Solve the following equation : 3  cos(tan–1x) = sin  cot 1  4 

[C.B.S.E. 2013]

ANSWER KEY EXERICISE – 1 (UNSOLVED PROBLEMS) 1. (i)

11. 0

3 5  2 (ii) (4 – 10) (iii)  2  3   –1

13. S = tan

  nx    1  (n  1)x 2   

8. x =

1 6

16. x =

9. x =

1 ,y=1 2

1 3

EXERICISE – 2 (BOARD PROBLEMS) Q2. 

Q3.

 4

Q4.

5 6

Q6.

 5

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10. x = ±1, ± (1 ±

2)