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(B) Given $\frac{\sin ^{-1} x}{\alpha}=\frac{\cos ^{-1} x}{\beta} = k$.We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$.Substituting the val

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

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(B) Given $\frac{\sin ^{-1} x}{\alpha}=\frac{\cos ^{-1} x}{\beta} = k$.We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$.Substituting the values,$k \alpha + k \beta = \frac{\pi}{2}$,which implies $k(\alpha + \beta) = \frac{\pi}{2}$,so $k = \frac{\pi}{2(\alpha + \beta)}$.From $\sin ^{-1} x = k \alpha$,we have $\alpha = \frac{\sin ^{-1} x}{k} = \frac{2(\alpha + \beta) \sin ^{-1} x}{\pi}$.Therefore,$\frac{\alpha}{\alpha + \beta} = \frac{2 \sin ^{-1} x}{\pi}$.We need to find $\sin \left(\frac{2 \pi \alpha}{\alpha + \beta}ight)$.Substituting the ratio,$\sin \left(2 \pi \cdot \frac{2 \sin ^{-1} x}{\pi}ight) = \sin (4 \sin ^{-1} x)$.Using the formula $\sin(4	heta) = 2 \sin(2	heta) \cos(2	heta) = 2(2 \sin 	heta \cos 	heta)(1 - 2 \sin^2 	heta) = 4 \sin 	heta \cos 	heta (1 - 2 \sin^2 	heta)$.Let $	heta = \sin ^{-1} x$,then $\sin 	heta = x$ and $\cos 	heta = \sqrt{1 - x^2}$.Substituting these,we get $4 x \sqrt{1 - x^2} (1 - 2 x^2)$.

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 $......$

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