Solve: sin^{- 1} (1 - x) - 2 sin^{- 1} x = π/2, then x is equal to:
(A) 0, 1/2   (B) 1, 1/2   (C) 0   (D) 1/2

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} (1 - x) - 2 sin^{- 1} x = π/2

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

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

⇒ - 2 sin^{- 1} x = cos^{- 1} (1 - x) ....(1)

Let sin^{- 1} x = y ⇒ sin y = x

Hence,

cos y = √1 - x²

y = cos^-1 (√1 - x²)

sin^{- 1} x = cos^{- 1} (√1 - x²)

From equation (1), we have

- 2 cos^{- 1} (√1 - x²) = cos^{- 1} (1 - x)

Put x = sin y

⇒ - 2 cos^{- 1} √1 - sin² y = cos^{- 1} (1- sin y)

⇒ - 2 cos^{- 1} (cos y ) = cos^{- 1} (1 - sin y)

⇒ - 2 y = cos^{- 1} (1 - sin y)

⇒ 1 - sin y = cos(- 2y)

⇒ 1 - sin y = cos 2y

⇒ 1 - sin y = 1 - 2 sin² y

⇒ 2 sin² y - sin y = 0

⇒ sin y (2 sin y - 1) = 0

⇒ sin y = 0, 1/2

Therefore,

x = 0, 1/2

When x = 1/2, it does not satisfy the equation.

Hence, x = 0 is the only solution

Thus, the correct option is C

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NCERT Solutions for Class 12 Maths - Chapter 2 Exercise ME Question 16

Solve: sin^{- 1} (1 - x) - 2 sin^{- 1} x = π/2, then x is equal to: (A) 0, 1/2 (B) 1, 1/2 (C) 0 (D) 1/2

Summary:

For the function: sin^{- 1} (1 - x) - 2 sin^{- 1} x = π/2, the value of x is 0.Thus, the correct option is C