Find the value of sin^{- 1} x = y, then
(A) 0 ≤ y ≤ π
(B) - π/2 ≤ y ≤ π/2
(C) 0 ≤ y ≤ p
(D) - π/2 < y < π/2

Solution:

Inverse trigonometric functions are the inverse ratio of the basic trigonometric ratios

Here the basic trigonometric function of Sin θ = y can be changed to θ = sin^-1 y

It is given that

sin^{- 1} x = y

Range of the principal value of sin^{- 1} x

= [- π / 2, π / 2]

Thus,

- π / 2 ≤ y ≤ π / 2

The answer is B

NCERT Solutions for Class 12 Maths - Chapter 2 Exercise 2.1 Question 13

Find the value of sin^{- 1} x = y, then (A) 0 ≤ y ≤ π (B) - π/2 ≤ y ≤ π/2 (C) 0 ≤ y ≤ p (D) - π/2 < y < π/2

Summary:

The value of sin^{- 1} x = y,  is - π / 2 ≤ y ≤ π / 2. The answer is B. Range of the principal value of sin^{- 1} x = [- π / 2, π / 2]