#### NCERT Solutions for Class 12th: Ch 2 Inverse Trigonometric Functions Exercise 2.2 Math

Page No: 41

**Exercise 2.1**

Prove the following:

1. 3sin

1. 3sin

^{-1}x = sin^{-1}(3x – 4x3), x ∈ [-/2, 1/2]**Answer**

To prove:

3sin

^{-1}x = sin^{-1}(3x − 4x3), x ∈ [−1/2, 1/2]
Let sin

^{-1}x = Î¸, then x = sin Î¸.
We have,

RHS = sin-1(3x - 4x

^{3})
= sin

^{-1}(3 sin Î¸ - 4sin3Î¸)
= sin

^{-1}(sin 3Î¸) = 3Î¸
= 3sin

^{-1}x = LHS
2. 3cos

^{-1}x = cos^{-1}(4x^{3}– 3x) x ∈ [1, 1/2]**Answer**

To prove:

3cos

^{-1}x = cos^{-1}(4x^{3}– 3x) x ∈ [1, 1/2].
Let cos

^{-1}x = Î¸, then x = cos Î¸.
We have,

RHS = cos

^{-1}(4x^{3}- 3x)
= cos

^{-1}(4cos^{3}Î¸ - 3cosÎ¸)
= cos

^{-1}(cos 3Î¸) = 3Î¸
= 3cos

^{-1}x
= LHS

3. tan

^{-1}2/11 + tan^{-1}7/24 = tan^{-1}1/2**Answer**

To prove: tan

LHS = tan

Exercise 2.1 Inverse Trigonometry

^{-1}2/11 + tan^{-1}7/24 = tan^{-1}1/2LHS = tan

^{-1}2/11 + tan^{-1}7/24
= tan

^{-1}(48 + 77)/(264 − 14)
= tan

^{-1}125/250 = tan^{-1}1/2 = RHS
4. 2tan

^{-1}1/2 + tan^{-1}1/7 = tan^{-1}31/17**Answer**

To prove: 2tan

^{-1}1/2 + tan^{-1}1/7 = tan^{-1}31/17
LHS = 2tan-11/2 + tan-11/7

Write the following functions in the simplest form:

Question: 5

**Answer**

Question: 6

**Answer**

Question: 7

**Answer**

Question: 8

**Answer**

Page No. 48

Question: 9

Answer

Question: 10

**Answer**

Find the values of each of the following:

Question: 11

**Answer**

Question: 12. cot (tan

^{-1}a + cot^{-1}a)**Answer**

The given function is cot(tan

^{-1}a + cot^{-1}a).
∴ cot(tan

^{-1}a + cot^{-1}a)
= cot (Ï€/2) [tan

^{-1}x + cot^{-1}x = Ï€/2]
= 0

Question: 13

**Answer**

Formula used:

Question: 14

**Answer**

Question: 15

**Answer**

Question: 16

**Answer**

Question: 17

**Answer**

Question: 18

**Answer**

Question: 19

**Answer**

The correct option is B.

Question: 20

**Answer**

The correct option is D.

Question: 21

**Answer**

The correct option is B.