If sin ^{-1}left(frac{x}{5}right) + csc ^{-1} | vedclass

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(D) Given the equation: $\sin ^{-1}\left(\frac{x}{5}\right) + \csc ^{-1}\left(\frac{5}{4}\right) = \frac{\pi}{2}$ We know that $\csc ^{-1}(z) = \sin ^

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$3$

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(D) Given the equation: $\sin ^{-1}\left(\frac{x}{5}ight) + \csc ^{-1}\left(\frac{5}{4}ight) = \frac{\pi}{2}$ We know that $\csc ^{-1}(z) = \sin ^{-1}\left(\frac{1}{z}ight)$. Therefore,$\csc ^{-1}\left(\frac{5}{4}ight) = \sin ^{-1}\left(\frac{4}{5}ight)$. Substituting this into the equation: $\sin ^{-1}\left(\frac{x}{5}ight) + \sin ^{-1}\left(\frac{4}{5}ight) = \frac{\pi}{2}$ Rearranging the terms: $\sin ^{-1}\left(\frac{x}{5}ight) = \frac{\pi}{2} - \sin ^{-1}\left(\frac{4}{5}ight)$ Using the identity $\sin ^{-1}(y) + \cos ^{-1}(y) = \frac{\pi}{2}$,we have $\frac{\pi}{2} - \sin ^{-1}\left(\frac{4}{5}ight) = \cos ^{-1}\left(\frac{4}{5}ight)$. So,$\sin ^{-1}\left(\frac{x}{5}ight) = \cos ^{-1}\left(\frac{4}{5}ight)$. To solve for $x$,convert $\cos ^{-1}\left(\frac{4}{5}ight)$ to $\sin ^{-1}$. Since $\cos 	heta = \frac{4}{5}$,the opposite side is $\sqrt{5^2 - 4^2} = 3$,so $\sin 	heta = \frac{3}{5}$. Thus,$\sin ^{-1}\left(\frac{x}{5}ight) = \sin ^{-1}\left(\frac{3}{5}ight)$. Comparing the arguments,$\frac{x}{5} = \frac{3}{5}$,which gives $x = 3$.

If $\sin ^{-1}\left(\frac{x}{5}\right) + \csc ^{-1}\left(\frac{5}{4}\right) = \frac{\pi}{2}$,then $x = $

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