NCERT Solutions for Class 12-science Maths Chapter 2 - Inverse Trigonometric Functions

Page / Exercise

Chapter 2 - Inverse Trigonometric Functions Ex. 2.1

Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7

 

We know that the range of the principal value of sec-1 is 

Solution 8

Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14

Chapter 2 - Inverse Trigonometric Functions Ex. 2.2

Solution 1

Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14

sin (sin-1 (1/5) + cos-1x) = 1

∴ sin-1 (1/5) + cos-1 x = sin-1 1

sin-1 (1/5) + cos-1 x = π/2

sin-1 (1/5) = π/2 - cos-1 x

sin-1 (1/5) = sin-1 x

∴ x = 1/5

We use the results: sin-1 1 = π/2 and sin-1 x + cos-1 x = π/2

Concept Insight:

As R.H.S is ‘1’, it is easier to take sin-1 1 = π/2

Solution 15

Solution 16
Solution 17
Solution 18

Solution 19
Solution 20
Solution 21

Chapter 2 - Inverse Trigonometric Functions Misc. Ex.

Solution 1
Solution 2
Solution 3
Solution 4

Solution 5

                  = begin mathsize 12px style cos to the power of negative 1 end exponent 33 over 65 end style                       by (3)

              = RHS

 

Solution 6

Let sin-1(3/5) = A and cos-1 (12/13) = B

So sin A = 3/5 and cos B = 12/13

Hence cos A = 4/5 and sin B = 5/13

As R.H.S is sin-1 we use sin (A + B)

Sin (A + B) = sin A cos B + cos A sin B = (3/5) (12/13) + (4/5) (5/13)

                                                                       = 36/65 + 20/65 = 56/65

Thus A + B = sin-1 (56/65) hence proved.

Concept insight:

If R.H.S is cos-1 or sin-1 then use Cos (A + B) or sin (A + B) as the case may be.

Solution 7

Solution 8

Solution 9

 

Solution 10

Solution 11
Solution 12

Solution 13

Solution 14

Solution 15

Solution 16


Solution 17