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

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## 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  = 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 ### STUDY RESOURCES

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