If $0<x<\frac{1}{\sqrt{2}}$ and $\frac{\sin ^{-1} x}{\alpha}=\frac{\cos ^{-1} x}{\beta}$, then a value of $\sin \left(\frac{2 \pi \alpha}{\alpha+\beta}\right)$ is

(B) $4 x \sqrt{\left(1-x^{2}\right)}\left(1-2 x^{2}\right)$

(A) $4 \sqrt{\left(1-x^{2}\right)}\left(1-2 x^{2}\right)$

(C) $2 x \sqrt{\left(1-x^{2}\right)}\left(1-4 x^{2}\right)$

(D) $4 \sqrt{\left(1-x^{2}\right)}\left(1-4 x^{2}\right)$

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Inverse Trigonometric Functions

0 < x < \frac{1}{2}

and

\frac{s i n ^{- 1} x}{α} = \frac{c o s ^{- 1} x}{β}

, then a value of

sin (\frac{2 π α}{α + β})

Answer: (B)

Let

\frac{s i n ^{- 1} x}{α} = \frac{c o s ^{- 1} x}{β} = k

. Then

sin^{- 1} x = k α

and

cos^{- 1} x = k β

. Since

sin^{- 1} x + cos^{- 1} x = \frac{π}{2}

, we have

k (α + β) = \frac{π}{2}

, so

k = \frac{π}{2 ( α + β )}

. Thus

sin^{- 1} x = \frac{π α}{2 ( α + β )}

, and

\frac{2 π α}{α + β} = 4 sin^{- 1} x

. Now

sin (4 sin^{- 1} x) = 2 sin (2 sin^{- 1} x) cos (2 sin^{- 1} x) = 2 \cdot 2 x 1 - x^{2} \cdot (1 - 2 x^{2}) = 4 x 1 - x^{2} (1 - 2 x^{2})

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Single correct option