If 0<x<12 and sin−1⁡xα=cos−1⁡xβ, then a value of sin⁡2παα+β is

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If $0 < x < \frac{1}{\sqrt{2}}$ and $\frac{\sin^{- 1} x}{α} = \frac{\cos^{- 1} x}{β}$ , then a value of $\sin (\frac{2 πα}{α + β})$ is

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

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

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

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

answer is B.

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Detailed Solution

$\frac{\sin^{- 1} x}{α} = \frac{\cos^{- 1} x}{β} = k$

$\begin{array}{l} \sin^{- 1} x = kα \\ \cos^{- 1} x = kβ \\ k = \frac{π}{2 (α + β)} . . . (i) \\ \sin (\frac{2 πα}{α + β}) = \sin (4 \sin^{- 1} x) \\ = 2 \sin (2 \sin^{- 1} x) \cos (2 \sin^{- 1} x) \\ = 4 x \sqrt{1 - x^{2}} (1 - 2 x^{2}) \end{array}$