If sin ^{-1} frac{1}{3}+sin ^{-1} frac{3}{5}+ | vedclass

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(B) Given the equation: $\sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{3}{5}+\sin ^{-1} x=\frac{\pi}{2}$.We know that $\sin ^{-1} x = \cos ^{-1} \sqrt{1-x^2

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$\frac{8 \sqrt{2}-3}{15}$

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(B) Given the equation: $\sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{3}{5}+\sin ^{-1} x=\frac{\pi}{2}$.We know that $\sin ^{-1} x = \cos ^{-1} \sqrt{1-x^2}$.So,$\sin ^{-1} \frac{1}{3} = \cos ^{-1} \sqrt{1-(\frac{1}{3})^2} = \cos ^{-1} \sqrt{\frac{8}{9}} = \cos ^{-1} \frac{2\sqrt{2}}{3}$.And $\sin ^{-1} \frac{3}{5} = \cos ^{-1} \sqrt{1-(\frac{3}{5})^2} = \cos ^{-1} \sqrt{\frac{16}{25}} = \cos ^{-1} \frac{4}{5}$.Substituting these into the equation: $\cos ^{-1} \frac{2\sqrt{2}}{3} + \cos ^{-1} \frac{4}{5} + \sin ^{-1} x = \frac{\pi}{2}$.$\sin ^{-1} x = \frac{\pi}{2} - (\cos ^{-1} \frac{2\sqrt{2}}{3} + \cos ^{-1} \frac{4}{5})$.Using $\frac{\pi}{2} - \cos ^{-1} 	heta = \sin ^{-1} 	heta$,we get $\sin ^{-1} x = \sin ^{-1} (\cos ^{-1} \frac{2\sqrt{2}}{3} + \cos ^{-1} \frac{4}{5})$.Let $\alpha = \cos ^{-1} \frac{2\sqrt{2}}{3}$ and $\beta = \cos ^{-1} \frac{4}{5}$. Then $\cos \alpha = \frac{2\sqrt{2}}{3}$ and $\cos \beta = \frac{4}{5}$.Then $\sin \alpha = \sqrt{1-(\frac{2\sqrt{2}}{3})^2} = \frac{1}{3}$ and $\sin \beta = \sqrt{1-(\frac{4}{5})^2} = \frac{3}{5}$.We need $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta = (\frac{1}{3} 	imes \frac{4}{5}) + (\frac{2\sqrt{2}}{3} 	imes \frac{3}{5}) = \frac{4}{15} + \frac{6\sqrt{2}}{15} = \frac{4+6\sqrt{2}}{15}$.Wait,re-evaluating: $\sin ^{-1} x = \frac{\pi}{2} - (\sin ^{-1} \frac{1}{3} + \sin ^{-1} \frac{3}{5}) = \cos ^{-1} (\sin ^{-1} \frac{1}{3} + \sin ^{-1} \frac{3}{5})$.Let $A = \sin ^{-1} \frac{1}{3}$ and $B = \sin ^{-1} \frac{3}{5}$. Then $\sin A = \frac{1}{3}, \cos A = \frac{2\sqrt{2}}{3}$ and $\sin B = \frac{3}{5}, \cos B = \frac{4}{5}$.$\sin(A+B) = \sin A \cos B + \cos A \sin B = (\frac{1}{3} 	imes \frac{4}{5}) + (\frac{2\sqrt{2}}{3} 	imes \frac{3}{5}) = \frac{4+6\sqrt{2}}{15}$.Since $\sin ^{-1} x = \frac{\pi}{2} - (A+B)$,then $x = \sin(\frac{\pi}{2} - (A+B)) = \cos(A+B) = \cos A \cos B - \sin A \sin B$.$x = (\frac{2\sqrt{2}}{3} 	imes \frac{4}{5}) - (\frac{1}{3} 	imes \frac{3}{5}) = \frac{8\sqrt{2}-3}{15}$.Thus,the correct option is $B$.

If $\sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{3}{5}+\sin ^{-1} x=\frac{\pi}{2}$,then $x=$

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