NCERT Solution Class 12 Maths by abhishek.sangwan1495

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Class XII – NCERT – Maths

Chapter 2- Inverse trigonometric functions

Exercise 2.1 Question 1:  1 Find the principal value of sin 1     2 Solution 1:  1      1 Let sin 1     y, Then sin y       sin    sin    .  2 6  6  2

   We know that the range of the principal value branch of sin 1 is   ,   2 2   1 and sin     ,  6 2

  1 Therefore, the principal value of sin 1    is  . 6  2

Question 2:  3 Find the principal value of cos 1    2  Solution 2:  3 3    cos   Let cos 1    y . Then cos y  2 6  2 

We know that the range of the principal value branch of cos 1 is 0,   3   and cos    . 6 2  3  Therefore, the principal value of cos 1   is .  2  6

Question 3: Find the principal value of cosec1 (2) Solution 3:

2.Inverse trigonometric functions

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Study Materials NCERT Solutions for Class 6 to 12 (Math & Science) Revision Notes for Class 6 to 12 (Math & Science) RD Sharma Solutions for Class 6 to 12 Mathematics RS Aggarwal Solutions for Class 6, 7 & 10 Mathematics Important Questions for Class 6 to 12 (Math & Science) CBSE Sample Papers for Class 9, 10 & 12 (Math & Science) Important Formula for Class 6 to 12 Math CBSE Syllabus for Class 6 to 12 Lakhmir Singh Solutions for Class 9 & 10 Previous Year Question Paper CBSE Class 12 Previous Year Question Paper CBSE Class 10 Previous Year Question Paper JEE Main & Advanced Question Paper NEET Previous Year Question Paper

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Chapter 2- Inverse trigonometric functions

Class XII – NCERT – Maths

Let cosec1 (2)  y .   Then, cosec y = 2 = cosec   . 6    We know that the range of the principal value branch of cosec -1 is   ,   0 .  2 2

Therefore, the principal value of cosec

Question 4:



Find the principal value of tan 1  3

-1

 2  is

 . 6



Solution 4: Let tan 1  3  y





Then, tan y   3   tan



   tan    . 3  3

  We know that the range of the principal value branch of tan 1 is     2 2   and tan    is  3.  3





Therefore, known that the principal value of tan 1  3 is 

 . 3

Question 5:  1 Find the principal value of cos 1     2

Solution 5:  1 Let cos 1     y.  2 1      2 Then, cos y     cos    cos      cos  2 3 3   3

 .  We know that the range of the principal value branch of cos 1 is 0,    2 and cos   3

1   . 2 

2.Inverse trigonometric functions

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Class XII – NCERT – Maths

Chapter 2- Inverse trigonometric functions

 1   2  Therefore, the principal value of cos 1    is  .  2  3 

Question 6: Find the principal value of tan 1  1 Solution 6: Let tan 1  1 = y.     Then, tan y  1   tan    tan    . 4  4    We know that the range of the principal value branch of tan 1 is   ,   2 2   and tan     1.  4

Therefore, the principal value of tan 1  1 is 

 . 4

Question 7:  2  Find the principal value of sec 1    3

Solution 7: 2  2    Let sec 1   sec   .   y . Then, sec y  3 6  3

  We know that the range of the principal value branch of sec1 is  0,      2   2 and sec    . 3 6  2   Therefore, the principal value of sec 1   is .  3 6

2.Inverse trigonometric functions

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Class XII – NCERT – Maths

Question 8: Find the principal value of cot 1

Chapter 2- Inverse trigonometric functions

 3

Solution 8: Let cot 1

 3   y . Then cot y 

  3  cot   . 6

We know that the range of the principal value branch of cot 1 is (0, )   and cot    3. 6

Therefore, the principal value of cot 1

 3  is 6 .

Question 9:  1  Find the principal value of cos 1    2 

Solution 9:  1  Let cos 1    y . 2 

Then cos y  

1      3   cos    cos      cos  4 2 4   4

 . 

We know that the range of the principal value branch of cos 1 is [0, ] 1  3  and cos     . 2  4   1  3 Therefore, the principal value of cos 1    is 4 . 2 

Question 10:



Find the principal value of cosec-1  2



Solution 10:

2.Inverse trigonometric functions

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Class XII – NCERT – Maths



Chapter 2- Inverse trigonometric functions



  Let cosec-1  2  y. Then, cos ec y   2   cos ec    cos ec    . 4  4 We know that the range of the principal value branch of cosec-1 is        2 , 2   {0} and cosec   4    2 .





Therefore, the principal value of cosec-1  2 is 

 . 4

Question 11:  1  1 Find the value of tan 1 1  cos 1     sin 1     2  2

Solution 11: Let tan 1 1  x.   Then, tan x =1= tan   . 4

 tan 1 1 

 4

 1 Let cos 1     y.  2 1      2 Then, cos y     cos    cos      cos  2 3 3   3

 . 

 1  2  cos-1      2 3  1 Let sin 1     z.  2

Then, sin z  

1       sin    sin    . 2 6  6

  1  sin 1      6  2  1  1  tan 1 1  cos 1     sin 1     2  2

2.Inverse trigonometric functions

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Chapter 2- Inverse trigonometric functions

Class XII – NCERT – Maths



 4



2   3 6

3  8  2 9 3   12 12 4

Question 12: 1 1 Find the value of cos 1    2sin 1   2 2

Solution 12: 1 Let cos 1    x. 2 1   Then, cos x   cos   . 2 3 1   cos 1    2 3 1 Let sin 1    y. 2 1   Then, sin y   sin   . 2 6 1   sin 1    2 6    2 1 1   cos 1    2sin 1     2     6 3 3 3 2 2 3

Question 13: If sin 1 x  y, then (A) 0  y  

(B) 

(D) 

   y 2 2   2

 y

Solution 13:

2.Inverse trigonometric functions

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Class XII – NCERT – Maths

Chapter 2- Inverse trigonometric functions

It is given that sin 1 x  y .    We know that the range of the principal value branch of sin 1 is   ,  .  2 2

Therefore, 

   y . 2 2

Answer choice (B) is correct.

Question 14:

tan 1 3  sec1  2 is equal to (B)  / 3 (D) 2 / 3

(A)  (C)  / 3

Solution 14: Let tan 1 3  x. . Then, tan x  3  tan

 3

    We know that the range of the principal value branch of tan 1 is  , .  2 2

 3 1 Let sec  2  y.

 tan 1 3 

 2    Then, sec y  2   sec    sec      sec . 3 3 3    We know that the range of the principal value branch of sec1 is  0,     . 2 2  sec 1  2   3 Thus,

tan 1 

 3

 3   sec



1

 2 

2   3 3

2.Inverse trigonometric functions

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Class XII – NCERT – Maths

Chapter 2- Inverse trigonometric functions Exercise 2.2

Question 1:  1 1 Prove 3sin 1 x  sin 1  3 x  4 x3  , x    ,   2 2 Solution 1:

 1 1 To Prove 3sin 1 x  sin 1  3x  4 x3  , where x    ,   2 2

Let x  sin  . Then,sin 1 x   . We have, R.H.S

sin 1  3x  4 x3   sin 1  3sin   4sin 3  

 sin 1  sin 3   3  3sin 1 x  L.H .S

Question 2: 1  Prove 3cos 1 x  cos 1  4 x3  3x  , x   ,1 2 

Solution 2: 1  To Prove 3cos 1 x  cos 1  4 x3  3x  , x   ,1 2 

Let x  cos . Then, cos 1 x   We have R.H .S cos 1  4 x3  3 x 

2.Inverse trigonometric functions

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Class XII – NCERT – Maths

Chapter 2- Inverse trigonometric functions

 cos 1  4 cos3   3cos    cos 1  cos 3   3  3cos 1 x  L.H .S

Question 3: 1  2  1  7  1  1  Prove tan    tan    tan    11   24  2

Solution 3: 1 To prove: tan

2 7 1  tan 1  tan 1 11 24 2

L.H .S . tan 1

2 7  tan 1 11 24

2 7   tan 1 11 24 2 7  1  .   11 24   tan 1

 tan 1

 1 1 1 x  y   tan x  tan y  tan 1  xy   

11 24 11 24  14 11 24 48  77 264  14

 125  1  1   tan 1    tan    R.H.S.  250  2

2.Inverse trigonometric functions

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Class XII – NCERT – Maths

Chapter 2- Inverse trigonometric functions

Question 4: 1 Prove 2 tan

1 1 31  tan 1  tan 1 2 7 17

Solution 4:

1 1 31  tan 1  tan 1 2 7 17 1 1 L.H.S = 2tan -1  tan 1 2 7 1 2. 2  tan 1 1  2 tan 1 x  tan 1 2 x   tan 1 2 7  1  x 2  1 1   2 1 1  tan 1  tan 1 7 3   4 4 1  tan 1  tan 1 3 7 4 1   x y   tan 1 3 7  tan 1 x  tan 1 y  tan 1 4 1 1  xy  1 .  3 7  28  3     tan 1  21  21  4    21  31  tan 1  R.H.S. 17

1 To prove: 2 tan

Question 5: Write the function in the simplest form: tan

1

1  x2 1 ,x 0 x

Solution 5:

tan 1

1  x2 1 x

Put x  tan     tan 1 x

2.Inverse trigonometric functions

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Class XII – NCERT – Maths

Chapter 2- Inverse trigonometric functions

1  x2  1 1  tan 2    tan 1 x tan   sec   1  1  1  cos    tan 1    tan    tan    sin  

 tan 1

 2   2sin  2      tan 1    2sin cos    2 2     1   tan 1  tan    tan 1 x 2 2 2 

Question 6:  1  Write the function in the simplest form: tan 1  2  , x  1  x 1 

Solution 6: 1 tan 1 2 , x  1 x 1

Put x  cosec    cos ec1 x 1  tan 1 x2 1 1  tan 1 cos ec 2  1  1   tan 1    cot    tan 1  tan  

  cos ec 1 x 

 2

 sec1 x

  1 1  As, cos ec x  sec x  2 

2.Inverse trigonometric functions

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Class XII – NCERT – Maths

Chapter 2- Inverse trigonometric functions

Question 7:  1  cos x  Write the function in the simplest form: tan 1   , x    1  cos x 

Solution 7:  1  cos x  tan 1   , x    1  cos x 

 1  cos x  tan 1    1  cos x   2 x  2sin 2  tan 1   2 cos 2 x  2 

     

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



x 2

Question 8:  cos x  sin x  Write the function in the simplest form: tan 1  ,0  x    cos x  sin x 

Solution 8:  cos x  sin x  tan 1    cos x  sin x 

  sin x    1   cos x    1  tan    1   sin x     cos x     

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Chapter 2- Inverse trigonometric functions

Class XII – NCERT – Maths

 1  tan x   tan 1    1  tan x   1 x  y 1 1   tan 1  xy  tan x  tan y   

 tan 1 1  tan 1  tan x 

 x 4

Question 9: Write the function in the simplest form: tan 1

x a  x2 2

, x a

Solution 9: x tan 1 2 a  x2 Let, x  a sin  

 tan 1

x  x  sin   sin 1   a a

x a  x2 2

  a sin   tan 1   2 2 2  a  a sin  

 a sin    tan 1   2  a 1  sin    a sin    tan 1    a cos  

 tan 1  tan      sin 1

x a

Question 10:

 3a 2 x  x3  a a Write the function in the simplest form: tan 1  3 , a  0; x 2  3 3  a  3ax  Solution 10:

2.Inverse trigonometric functions

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Class XII – NCERT – Maths

Chapter 2- Inverse trigonometric functions

 3a 2 x  x3  Consider, tan  3 2   a  3ax  Let x  x x  a tan    tan     tan 1   a a 1

 3a 2 x  x3  tan 1  3 2   a  3ax   3a 2  a tan   a 3 tan 3    tan   3 2 2  a  3a  a tan   1

 3a3 tan   a3 tan 3    tan   3 3 2  a  3a tan   1

 tan 1  tan 3   3

 3 tan 1

x a

Question 11:  1   Find the value of tan 1  2 cos  2sin 1   2   

Solution 11: 1 Let sin 1  x. 2 1   Then,sin x   sin   . 2 6

 sin 1

1   2 6

 1    tan 1  2 cos  2sin 1   2         tan 1  2 cos  2    6   

2.Inverse trigonometric functions

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Chapter 2- Inverse trigonometric functions

Class XII – NCERT – Maths

   tan 1  2 cos  3   1  tan 1  2    2  tan 1 1 

 4

Question 12:

Find the value of cot  tan 1 a  cot 1 a  Solution 12:

cot  tan 1 a  cot 1 a     cot   2

  1 1  tan x  cot x  2 

0

Question 13: 2 1  1 2 x 1 1  y  Find the value of tan sin  cos , x  1, y  0 and xy  1 2 1  x2 1  y 2 

Solution 13: Let x  tan  . Then,  tan 1 x.

 sin 1

2x  2 tan    sin 1   2 2 1 x  1  tan    sin 1  sin 2   2  2 tan 1 x

Let y  tan  . Then,   tan 1 y.

2.Inverse trigonometric functions

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Class XII – NCERT – Maths

Chapter 2- Inverse trigonometric functions

 1 y2   cos  2   1 y   1  tan 2    cos 1   2  1  tan    cos 1  cos 2  1

 2  2 tan 1 y 2  1  1  2 x  1  1  y  cos  tan sin      2 2  2  1 x   1  y 

 1 1 1 x  y   As, tan x  tan y  tan 1  xy   

1  tan  2 tan 1 x  2 tan 1 y  2   x  y   tan  tan 1    1  xy    x y  1  xy

Question 14: 1   If sin  sin 1  cos 1 x   1 , then find the value of x. 5   Solution 14: 1   sin  sin 1  cos 1 x   1 5  

1 1    sin  sin 1  cos  cos 1 x   cos  sin 1  sin  cos 1 x   1 5 5    sin  A  B   sin A  cos B  cos A  sin B  1 1    x  cos  sin 1  sin  cos 1 x   1 5 5  

1 x   cos  sin 1  sin  cos 1 x   1 5 5 

Now, let sin 1

...(1)

1 y 5

Then,

2.Inverse trigonometric functions

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Chapter 2- Inverse trigonometric functions

Class XII – NCERT – Maths

1 y 5 1 sin y  5 sin 1

2 6 1  cos y  1     5 5 2 6  y  cos 1    5   sin 1

2 6 1  cos 1   5  5 

...(2)

Let cos1 x  z. Then, cos z  x  sin z  1  x 2  z  sin 1



1  x2

 cos 1 x  sin 1





1  x2



...(3)

From (1), (2) and (3) we have:





 x 2 6 1 2  cos  cos 1   sin sin 1  x  1 5 5   

x 2 6   1  x2  1 5 5

 x  2 6 1  x2  5  2 6 1  x2  5  x

On squaring both sides, we get:

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Chapter 2- Inverse trigonometric functions

Class XII – NCERT – Maths

 2 6  1  x   25  x 2

 10 x

  4  6  1  x 2   25  x 2  10 x  24  24 x 2  25  x 2  10 x  25 x 2  10 x  1  0   5 x  1  0 2

  5 x  1  0 x

1 5

Hence, the value of x is

1 . 5

Question 15: x 1  1 x  1  tan 1  , then find the value of x. If tan x2 x2 4 Solution 15: x 1 x 1  tan 1  tan 1  x2 x2 4

 x 1 x 1     1 x  2 x2     tan  1   x  1  x  1   4    x  2   x  2  

 1 1 1 x  y   As, tan x  tan y  tan 1  xy   

  x  1 x  2    x  1 x  2     tan 1     x  2  x  2    x  1 ( x  1)  4

 x2  x  2  x2  x  2    tan 1   x 2  4  x 2  1  4   2x2  4    tan 1    3  4

 4  2 x2    tan  tan 1  tan  3  4  

4  2 x2 1 3

2.Inverse trigonometric functions

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Chapter 2- Inverse trigonometric functions

Class XII – NCERT – Maths

 4  2 x2  3  2 x2  4  3  1 x

1 2

Hence, the value of x is 

1 . 2

Question 16: 2   Find the values of sin 1  sin  3  

Solution 16: 2   Consider, sin 1  sin  3  

We know that sin 1  sin x   x    If x    ,  , which is the principal value branch of sin 1 x. .  2 2

Here,

2      , 3  2 2 

2  Now, sin 1  sin 3 

  can be written as: 

2   sin 1  sin  3     2    sin 1 sin     3             sin 1  sin  , where   , 3 3  2 2   2   sin 1  sin 3 

   1    sin sin   3 3  

2.Inverse trigonometric functions

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Chapter 2- Inverse trigonometric functions

Class XII – NCERT – Maths

Question 17: 3   Find the values of tan 1  tan  4  

Solution 17: 3   Consider, tan 1  tan  4  

We know that tan 1  tan x   x    If x    ,  , which is the principal value branch of tan 1 x.  2 2 3     Here,    , . 4  2 2 3  Now, tan 1  tan 4  3  tan 1  tan 4 

  can be written as: 

    3    1  1    tan   tan     tan   tan      4    4    

           tan 1   tan   tan 1  tan    where     ,  4 4  2 2    4  3        1   tan 1  tan   tan  tan    4     4  4

Question 18: 3 3  Find the values of tan  sin 1  cot 1  5 2 

Solution 18: 1 3  x. Let sin 5 Then , 3 sin x  5

 cos x  1  sin 2 x   sec x 

4 5

5 4

2.Inverse trigonometric functions

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Chapter 2- Inverse trigonometric functions

Class XII – NCERT – Maths

 tan x  sec2 x  1 

 x  tan 1

25 3 1  16 4

3 4

3 3  sin 1  tan 1 5 4 2 1 3  tan 1 Now, cot 2 3

...(i)  1 1 1   , tan x  cot x 

...(ii)

3 3  Therefore, tan  sin 1  cot 1  5 2  3 2   tan  tan 1  tan 1  4 3 

[Using (i) and (ii)]

  3 2   1  4  3    tan  tan  3 2    1      4 3  

 1 1 1 x  y   As, tan x  tan y  tan 1  xy   

98    tan  tan 1  12  6   17  17   tan  tan 1   6 6 

Question 19: 7  Find the values of cos 1  cos 6 

( A) (C )

7 6

 3

  is equal to 

( B) ( D)

5 6

 6

Solution 19: We know that cos1  cos x   x if x 0,   , which is the principal value branch of cos1 x . Here,

7   0,   . 6

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7  Now, cos 1  cos 6  7  cos 1  cos 6 

  can be written as : 

7  1    cos  cos 6  

7   cos1  cos 6 

Chapter 2- Inverse trigonometric functions

5  1    cos  cos 6  

7     1    cos cos  2   6     

 cos  2  x   cos x 

 5   6

The correct answer is B.

Question 20:   1  Find the values of sin   sin 1     is equal to  2  3 1 1 (A) (B) 2 3 (D) 1 1 (C) 4

Solution 20:  1  Let sin 1    x .  2  Then , sin x 

1      sin  sin  2 6  6

 . 

    We know that the range of the principal value branch of sin 1 is  , .  2 2   1   sin 1     2  6   1       3  sin   sin 1     sin     sin   2  3 6  6 3 The correct answer is D.

2.Inverse trigonometric functions

     sin    1  2

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Chapter 2- Inverse trigonometric functions

Class XII – NCERT – Maths

Miscellaneous Exercise Question 1: 13   Find the value of cos 1  cos  6   Solution 1:

We Know that cos1  cos x   x if x 0,   , which is the principal value branch of cos1 x .

13   0,   . 6 13   Now, cos 1  cos  can be written as : 6   13   cos 1  cos  6        cos 1 cos  2    6    Here,

     cos 1 cos      6  13   cos 1  cos 6 

where

 6

  0,   .

     1    cos cos        6  6

Question 2: 7   Find the value of tan 1  tan  6  

Solution 2:    We know that tan 1  tan x   x if x    ,  , which is the principal value branch of tan 1 x.  2 2 7     Here,    , . 6  2 2 7  Now, tan 1  tan 6 

  can be written as: 

2.Inverse trigonometric functions

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Chapter 2- Inverse trigonometric functions

Class XII – NCERT – Maths

7   tan 1  tan  6     5  tan  tan  2  6   1

  5    tan 1   tan    6     5  tan 1  tan    6  

 tan  2  x    tan x 

  

 tan  2  x    tan x    

         tan 1  tan    , where    ,  6  2 2   6 

7   tan 1  tan 6 

   1    tan  tan   6 6  

Question 3: 1 Prove 2sin

3 24  tan 1 5 7

Solution 3: 3 1 3  x. Then sin x  . Let sin 5 5 2

4 3  cos x  1     5 5

 tan x 

3 4

3 3 3  sin 1  tan 1 4 5 4 Now, we have: L.H.S 3 3 2sin 1  2 tan 1 5 4  3   2  2x   1  4  2 tan 1 x  tan 1  tan 2   3  1  x 2    1     4 

 x  tan 1

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Chapter 2- Inverse trigonometric functions

 3     3 16   tan 1  2   tan 1    16  9 2 7     16  24  tan 1  R.H .S . 7

Question 4: 1 Prove sin

8 3 77  sin 1  tan 1 17 5 36

Solution 4: 8 sin 1  x. 17 2

8 225 15  8 Then, sin x   cos x  1      17 289 17.  17 

8 8  x  tan 1 15 15 8 8  sin 1  tan 1 17 15 3 Now, let sin 1  y 5  tan x 

…..(1)

3 16 4 3 Then, sin y   cos y  1      . 5 25 5 5

3 3  y  tan 1 5 4 3 3  sin 1  tan 1 5 4

 tan y 

……(2)

Now, we have: L.H.S.

sin 1

8 3  sin 1 17 5

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Using (1) and (2), we get 3 1 8  tan 1 = tan 15 4 8 3  1 15 4  tan 8 3 1  15 4  32  45   tan 1    60  24  77  tan 1  R.H .S . 36

Chapter 2- Inverse trigonometric functions

 1 1 1 x  y   tan x  tan y  tan 1  xy   

Question 5: 1 Prove cos

4 12 33  cos 1  cos 1 5 13 65

Solution 5: 4 Let cos 1  x 5 2

Then, cos x 

4 3 4  sin x  1     5 5 5

3 3  x  tan 1 4 4 4 3  cos 1  tan 1 ….(1) 5 4 1 12  y. Now, let cos 13 12 5 Then, cos y   sin y  . 13 13 5 5  tan y   y  tan 1 12 12 12 5  cos 1  tan 1 ….(2) 13 12  tan x 

1 Let cos

33 z 65

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Class XII – NCERT – Maths

33 56  sin z  65 65 56 56  tan z   z  tan 1 33 33 33 56  cos 1  tan 1 65 33

Chapter 2- Inverse trigonometric functions

Then, cos z 

...(3)

Now, L.H .S .

4 12 cos 1  cos 1 5 13 Using (1) and (2), we get

3 5  tan 1  tan 1 4 12  3 5      tan 1  4 12   3 5   1   4  12     

 1 1 1  x  y    tan x  tan y  tan    1  xy   

 36  20   tan 1    48  15  56  tan 1 33 56  cos 1 33  R.H .S .

Question 6: 12 3 56 Prove cos 1  sin 1  sin 1 13 5 65 Solution 6: 2

3 3 16 4 3  x. Then, sin x   cos x  1     Let sin  . 5 5 25 5 5 1

2.Inverse trigonometric functions

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Chapter 2- Inverse trigonometric functions

Class XII – NCERT – Maths

3 3  x  tan 1 4 4 3 3  sin 1  tan 1 ...(1) 5 4 12 Now, let cos 1  y 13

 tan x 

Then, cos y 

 tan y 

 cos 1

5 5  y  tan 1 12 12

12 5  tan 1 13 12

Let sin 1

 tan z 

...(2)

56  z. 65

Then, sin z 

 sin 1

12 5  sin y  . 13 13

56 33  cos z  . 65 65

56 56  z  tan 1 33 33

56 56  tan 1 65 33

...(3)

Now, we have L.H.S 12 3 cos 1  sin 1 13 5 5 3  tan 1  tan 1 12 4 Using (1) and (2), we get

2.Inverse trigonometric functions

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Class XII – NCERT – Maths

5 3  = tan -1 12 4  5 3 1  .   12 4 

Chapter 2- Inverse trigonometric functions

 1 1 1 x  y   tan x  tan y  tan 1  xy   

 20  36  = tan 1    48  15  56 = tan 1 33 56  sin 1 [Using (3)] 65

 R.H .S

Question 7: 63 5 3  sin 1  cos 1 Prove tan 1 16 13 5 Solution 7: 5 Let sin 1  x 13 5 12 Then, sin x   cos x  . 13 13 5 5  tan x   x  tan 1 12 12 5 5  sin _1  tan 1 13 12 3 Let cos 1  y. 5 3 4 Then, cos y   sin y  . 5 5 4 4 Thus, tan y   y  tan 1 3 3 3 4  cos 1  tan 1 5 3

…..(1)

….(2)

Using (1) and (2), we have R.H.S.

2.Inverse trigonometric functions

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Class XII – NCERT – Maths

sin 1

Chapter 2- Inverse trigonometric functions

5 3  cos 1 12 5

= tan 1

5 4  tan 1 12 3

 5 4   12  3  1  tan    1 5  4   12 3 

 1 1 1 x  y   tan x  tan y  tan 1  xy   

 15  48   tan 1    36  20  63  tan 1 16  L.H.S .

Question 8: 1 1 1 1  Prove tan 1  tan 1  tan 1  tan 1  5 7 3 8 4 Solution 8: L.H.S 1 1 1 1 tan 1  tan 1  tan 1  tan 1 5 7 3 8  1 1   1 1      38  1 1 5 7  tan    tan  1 1   1 1  1   1    5 7  3 8  75  1  8  3   tan 1    tan    35  1   24  1  12 11  tan 1  tan 1 34 23 6 11  tan 1  tan 1 17 23  6 11      tan 1  17 23   1  6  11   17 23 

2.Inverse trigonometric functions

 1 1 1 x  y   tan x  tan y  tan 1  xy   

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Class XII – NCERT – Maths

Chapter 2- Inverse trigonometric functions

 325  1  tan 1    tan 1 325      R.H.S. 4

Question 9: 1  1 x  Prove tan 1 x  cos 1   , x   0.1 2  1 x 

Solution 9: Let x  tan 2  . Then

x  tan     tan 1 x .

1  x 1  tan 2     cos 2 1  x 1  tan 2 

Now, we have: R.H .S 1  1 x  cos 1   2  1 x  1  cos 1  cos 2  2 1   2 =  =tan -1 x  L.H .S . 2

Question 10:  1  sin x  1  sin x  x   Prove cot 1    , x   0,    4  1  sin x  1  sin x  2

Solution 10:

2.Inverse trigonometric functions

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Chapter 2- Inverse trigonometric functions

Class XII – NCERT – Maths

Consider

    

1  sin x  1  sin x 1  sin x  1  sin x

 1  sin x 

1  sin x  1  sin x 1  sin x 

(by rationalizing)

1  sin x   1  sin x   2 1  sin x 1  sin x  1  sin x  1  sin x



2 1  1  sin 2 x



2sin x 1  cos x  sin x 

2 cos 2

x 2

x x 2sin cos 2 2 x  cot 2  1  sin x  1  sin x   L.H .S .  cot 1    1  sin x  1  sin x  x x   cot 1  cot    R.H .S . 2 2 

Question 11:  1 x  1 x   1 1 1 Prove tan 1   x 1    cos x,  2  1 x  1 x  4 2

Hint : putx  cos 2  Solution 11: Let, x  cos 2 then,  

1 cos 1 x. 2

Thus, we have:

2.Inverse trigonometric functions

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Class XII – NCERT – Maths

Chapter 2- Inverse trigonometric functions

 1 x  1 x  L.H .S .  tan 1    1 x  1 x   1  cos 2  1  cos 2  tan 1   1  cos 2  1  cos 2  2 cos 2 2  2sin 2   tan 1   2 cos 2 2  2sin 2  

  

   

 2 cos   2 sin    tan     2 cos   2 sin  

 cos   sin    tan 1    cos   sin    1  tan    tan 1    1  tan  

 1  x  y  1 1   tan    tan x  tan y   1  xy   

 tan 1 1  tan 1  tan   

 4

 



1  cos 1 x  R.H.S. 4 2

Question 12: Prove

9 9 1 1 9 1 2 2  sin  sin 8 4 3 4 3

Solution 12: 9 9 1 1 L.H.S. =  sin 8 4 3 9 1    sin 1  4 2 3 9 1   cos 1  4 3 1 Now, let cos 1  x 3

2.Inverse trigonometric functions

……(1)

  1 1  sin x  cos x  2 

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Class XII – NCERT – Maths

Chapter 2- Inverse trigonometric functions

Then, cos x 

1 2 2 1  sin x  1     . 3 3 3

 x  sin 1

2 2 1 2 2  cos 1  sin 1 3 3 3

 L.H.S. =

9 1 2 2 sin  R.H.S. 4 3

Question 13: Solve 2tan 1  cos x   tan 1  2cosec x  Solution 13:

2tan 1  cos x   tan 1  2cosec x   2cos x  1  tan 1    tan  2cosec x  2 1  cos x   2 cos x   2 cosec x 1  cos 2 x 

 1 1 2 x   2 tan x  tan 1  x 2 

2 cos x 2  2 sin x sin x

 cos x  sin x

 tan x 1 x 

 4

Question 14: 1 x 1  tan 1 x,  x  0  Solve tan 1 1 x 2 Solution 14: 1 x 1 tan 1  tan 1 x 1 x 2

2.Inverse trigonometric functions

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Chapter 2- Inverse trigonometric functions

Class XII – NCERT – Maths

 1 1 1 x  y   tan x  tan y  tan 1  xy   

1  tan 1 1  tan 1 x  tan 1 x 2 

 4



3 tan 1 x 2

 tan 1 x   x  tan

x 

 6

1 3

Question 15:

Solve sin  tan 1 x  , x  1 is equal to (A) (C)

(B)

1  x2 1

(D)

1  x2

1 1  x2 x 1  x2

Solution 15: tan y  x  sin y 

x 1  x2

Let tan 1 x  y. Then,

 x   x  y  sin 1   tan 1 x  sin 1    2 2  1 x   1 x    x  x  sin  tan 1 x   sin  sin 1      2  1  x2  1 x    The correct answer is D.

2.Inverse trigonometric functions

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Chapter 2- Inverse trigonometric functions

Class XII – NCERT – Maths

Question 16: Solve sin 1 1  x   2sin 1 x  (A) 0,

1 2



, then x is equal to

(B) 1,

(D)

Solution 16:

sin 1 1  x   2sin 1 x   2sin 1 x 



1 2

 2

 sin 1 1  x 

2  2sin x  cos 1 1  x  1

...(1)

Let sin 1 x    sin   x  cos   1  x 2 .    cos 1



1  x2

 sin 1 x  cos 1





1  x2



Therefore, from equation (1), we have 2cos 1





1  x 2  cos 1 1  x 

Let, x  sin y . Then, we have: 2cos 1





1  sin 2 y  cos 1 1  sin y 

 2cos1  cos y   cos1 1  sin y   2 y  cos1 1  sin y 

 1  sin y  cos  2 y   cos 2 y  1  sin y  1  2sin 2 y  2sin 2 y  sin y  0

 sin y  2sin y 1  0 1 2 1  x  0 OR x  2 1 When x  , it can be observed that: 2  sin y  0 or

2.Inverse trigonometric functions

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Class XII – NCERT – Maths

Chapter 2- Inverse trigonometric functions

1  1 L.H.S.  sin 1 1    sin 1 2  2 1 1  sin 1    2sin 1 2 2 1   sin 1 2      R.H .S . 6 2 1  x  is not a solution of the given equation. 2 Thus, x  0. Hence, the correct answer is C.

Question 17: x x y Solve tan 1    tan 1 is equal to x y  y   (A) (B) 2 3  3 (C) (D) 4 4 Solution 17: x  x y tan 1    tan 1    y  x y  x x y     y x y  1   tan   x  x  y   1       y  x  y    x x  y  y  x  y    y  x  y 1    tan  y  x  y  x x  y    y  x  y  

 1 1 1 x  y   tan y  tan y  tan 1  xy   

 x 2  xy  xy  y 2   tan 1   2 2  xy  y  x  xy 

2.Inverse trigonometric functions

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Class XII – NCERT – Maths

Chapter 2- Inverse trigonometric functions

 x2  y 2    tan  2  tan 1 1  2  4 x y  Hence, the correct answer is C. 1

2.Inverse trigonometric functions

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