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

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

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$3$

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(D) Given equation: $\sin ^{-1}\left(\frac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}$ We know that $\operatorname{cosec}^{-1}(y) = \sin ^{-1}\left(\frac{1}{y}\right)$. So,$\operatorname{cosec}^{-1}\left(\frac{5}{4}\right) = \sin ^{-1}\left(\frac{4}{5}\right)$. Substituting this into the equation: $\sin ^{-1}\left(\frac{x}{5}\right)+\sin ^{-1}\left(\frac{4}{5}\right)=\frac{\pi}{2}$ $\Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{4}{5}\right)$ Using the identity $\sin ^{-1}(z)+\cos ^{-1}(z)=\frac{\pi}{2}$,we get $\frac{\pi}{2}-\sin ^{-1}\left(\frac{4}{5}\right) = \cos ^{-1}\left(\frac{4}{5}\right)$. $\Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\cos ^{-1}\left(\frac{4}{5}\right)$ Now,convert $\cos ^{-1}\left(\frac{4}{5}\right)$ to $\sin ^{-1}$ form: $\cos ^{-1}\left(\frac{4}{5}\right) = \sin ^{-1}\left(\sqrt{1-\left(\frac{4}{5}\right)^2}\right) = \sin ^{-1}\left(\sqrt{1-\frac{16}{25}}\right) = \sin ^{-1}\left(\sqrt{\frac{9}{25}}\right) = \sin ^{-1}\left(\frac{3}{5}\right)$. $\Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\sin ^{-1}\left(\frac{3}{5}\right)$ Comparing both sides,we get $\frac{x}{5} = \frac{3}{5}$,which implies $x = 3$.

If $\sin ^{-1}\left(\frac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}$,then the value of $x$ is

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