If sec^{-1} x = csc^{-1} y,then cos^{-1} frac | vedclass

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(D) Given that $\sec^{-1} x = \csc^{-1} y$. We know that $\sec^{-1} x = \cos^{-1} \frac{1}{x}$ and $\csc^{-1} y = \sin^{-1} \frac{1}{y}$. Substituting

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$\frac{\pi}{2}$

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(D) Given that $\sec^{-1} x = \csc^{-1} y$. We know that $\sec^{-1} x = \cos^{-1} \frac{1}{x}$ and $\csc^{-1} y = \sin^{-1} \frac{1}{y}$. Substituting these into the given equation,we get $\cos^{-1} \frac{1}{x} = \sin^{-1} \frac{1}{y}$. We also know the identity $\sin^{-1} 	heta + \cos^{-1} 	heta = \frac{\pi}{2}$,which implies $\sin^{-1} \frac{1}{y} = \frac{\pi}{2} - \cos^{-1} \frac{1}{y}$. Substituting this into our equation: $\cos^{-1} \frac{1}{x} = \frac{\pi}{2} - \cos^{-1} \frac{1}{y}$. Rearranging the terms,we get $\cos^{-1} \frac{1}{x} + \cos^{-1} \frac{1}{y} = \frac{\pi}{2}$.

If $\sec^{-1} x = \csc^{-1} y$,then $\cos^{-1} \frac{1}{x} + \cos^{-1} \frac{1}{y} = $

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