If sin ^{-1} frac{x}{5}+sin ^{-1} frac{4}{5}= | vedclass

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(A) Given the equation: $\sin ^{-1} \frac{x}{5}+\sin ^{-1} \frac{4}{5}=\frac{\pi}{2}$ We know that $\sin ^{-1} y + \cos ^{-1} y = \frac{\pi}{2}$ for $

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$3$

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(A) Given the equation: $\sin ^{-1} \frac{x}{5}+\sin ^{-1} \frac{4}{5}=\frac{\pi}{2}$ We know that $\sin ^{-1} y + \cos ^{-1} y = \frac{\pi}{2}$ for $y \in [-1, 1]$. Thus,$\sin ^{-1} \frac{x}{5} = \frac{\pi}{2} - \sin ^{-1} \frac{4}{5}$. Using the identity,we get $\sin ^{-1} \frac{x}{5} = \cos ^{-1} \frac{4}{5}$. Let $\cos ^{-1} \frac{4}{5} = 	heta$,then $\cos 	heta = \frac{4}{5}$. Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin 	heta = \sqrt{1 - (\frac{4}{5})^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$. Therefore,$\sin ^{-1} \frac{x}{5} = \sin ^{-1} \frac{3}{5}$. Comparing both sides,we get $\frac{x}{5} = \frac{3}{5}$,which implies $x = 3$.

If $\sin ^{-1} \frac{x}{5}+\sin ^{-1} \frac{4}{5}=\frac{\pi}{2}$,then $x=$ . . . . . . .

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