If sin^{-1}(frac{x}{5}) + csc^{-1}(frac{5}{4} | vedclass

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(D) Given the equation: $\sin^{-1}(\frac{x}{5}) + \csc^{-1}(\frac{5}{4}) = \frac{\pi}{2}$ We know that $\csc^{-1}(\frac{1}{y}) = \sin^{-1}(y)$,so $\cs

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$3$

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(D) Given the equation: $\sin^{-1}(\frac{x}{5}) + \csc^{-1}(\frac{5}{4}) = \frac{\pi}{2}$ We know that $\csc^{-1}(\frac{1}{y}) = \sin^{-1}(y)$,so $\csc^{-1}(\frac{5}{4}) = \sin^{-1}(\frac{4}{5})$. Substituting this into the equation: $\sin^{-1}(\frac{x}{5}) + \sin^{-1}(\frac{4}{5}) = \frac{\pi}{2}$ Rearranging the terms: $\sin^{-1}(\frac{x}{5}) = \frac{\pi}{2} - \sin^{-1}(\frac{4}{5})$ Using the identity $\sin^{-1}(z) + \cos^{-1}(z) = \frac{\pi}{2}$,we have $\frac{\pi}{2} - \sin^{-1}(\frac{4}{5}) = \cos^{-1}(\frac{4}{5})$. Thus,$\sin^{-1}(\frac{x}{5}) = \cos^{-1}(\frac{4}{5})$. Since $\cos^{-1}(\frac{4}{5}) = \sin^{-1}(\sqrt{1 - (\frac{4}{5})^2}) = \sin^{-1}(\sqrt{1 - \frac{16}{25}}) = \sin^{-1}(\sqrt{\frac{9}{25}}) = \sin^{-1}(\frac{3}{5})$. Therefore,$\sin^{-1}(\frac{x}{5}) = \sin^{-1}(\frac{3}{5})$,which implies $\frac{x}{5} = \frac{3}{5}$,so $x = 3$.

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

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