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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$8$

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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 $\operatorname{cosec}^{-1}(y) = \sin ^{-1}\left(\frac{1}{y}\right)$ for $|y| \geq 1$. So, $\operatorname{cosec}^{-1}\left(\frac{5}{4}\right) = \sin ^{-1}\left(\frac{4}{5}\right)$. Substituting this into the equation: $\sin ^{-1}\left(\frac{x}{5}\right)+\sin ^{-1}\left(\frac{4}{5}\right)=\frac{\pi}{2}$. Using the identity $\sin ^{-1}(A) + \cos ^{-1}(A) = \frac{\pi}{2}$, we have $\sin ^{-1}\left(\frac{x}{5}\right) = \frac{\pi}{2} - \sin ^{-1}\left(\frac{4}{5}\right) = \cos ^{-1}\left(\frac{4}{5}\right)$. Since $\cos ^{-1}\left(\frac{4}{5}\right) = \sin ^{-1}\left(\sqrt{1 - (\frac{4}{5})^2}\right) = \sin ^{-1}\left(\sqrt{1 - \frac{16}{25}}\right) = \sin ^{-1}\left(\sqrt{\frac{9}{25}}\right) = \sin ^{-1}\left(\frac{3}{5}\right)$. Therefore, $\sin ^{-1}\left(\frac{x}{5}\right) = \sin ^{-1}\left(\frac{3}{5}\right)$, which implies $\frac{x}{5} = \frac{3}{5}$, so $x = 3$. Finally, $5 + x = 5 + 3 = 8$.

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

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