sin (π/3 - sin^{- 1} (- 1/2)) is equal to
(A) 1/2   (B) 1/3   (C) 1/4   (D) 1

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

Let sin^{- 1} (- 1 / 2) = x

Hence,

sin x = - 1 / 2

= - sin π / 6

= sin (- π / 6)

x = - π / 6 

Since, Range of principal value of sin^{- 1} (x)

= [- π / 2, π / 2]

Therefore,

sin^{- 1} (- 1/2) = - π / 6

Then,

sin (π / 3 - sin^{- 1} (- 1 / 2))

= sin (π / 3 + π / 6)

= sin π / 2

= 1

Thus, the correct option is D

---

NCERT Solutions for Class 12 Maths - Chapter 2 Exercise 2.2 Question 20

sin (π/3 - sin^{- 1} (- 1/2)) is equal to (A) 1/2 (B) 1/3 (C) 1/4 (D) 1

Summary:

sin (π/3 - sin^{- 1} (- 1/2)) is equal to 1. Inverse trigonometric functions are the inverse ratio of the basic trigonometric ratios