If $\left(\sin ^{-1} x\right)^{2}-\left(\cos ^{-1} x\right)^{2}=a, 0<x<1, a \neq 0$, then the value of $2 x^{2}-1$ is

(C) $\sin \left(\frac{2 a}{\pi}\right)$

(A) $\cos \left(\frac{2 a}{\pi}\right)$

(B) $\cos \left(\frac{4 \mathrm{a}}{\pi}\right)$

(D) $\sin \left(\frac{4 a}{\pi}\right)$

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Inverse Trigonometric Functions

(sin^{- 1} x)^{2} - (cos^{- 1} x)^{2} = a, 0 < x < 1, a \neq = 0

, then the value of

2 x^{2} - 1

Answer: (C)

Writing

θ = sin^{- 1} x

gives

cos^{- 1} x = \frac{π}{2} - θ

, so

(sin^{- 1} x)^{2} - (cos^{- 1} x)^{2} = (2 θ - \frac{π}{2}) \frac{π}{2} = a,

hence

2 θ = \frac{π}{2} + \frac{2 a}{π}

. Since

x = sin θ

2 x^{2} - 1 = - cos 2 θ = - cos (\frac{π}{2} + \frac{2 a}{π}) = sin (\frac{2 a}{π}) .

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Single correct option

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