If y = sec^{-1}left( frac{sqrt{x} + 1}{sqrt{x | vedclass

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(A) Given the expression: $y = \sec^{-1}\left( \frac{\sqrt{x} + 1}{\sqrt{x} - 1} \right) + \sin^{-1}\left( \frac{\sqrt{x} - 1}{\sqrt{x} + 1} \right)$

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(A) Given the expression: $y = \sec^{-1}\left( \frac{\sqrt{x} + 1}{\sqrt{x} - 1} ight) + \sin^{-1}\left( \frac{\sqrt{x} - 1}{\sqrt{x} + 1} ight)$ We know that $\sec^{-1}(z) = \cos^{-1}\left( \frac{1}{z} ight)$. Therefore,$\sec^{-1}\left( \frac{\sqrt{x} + 1}{\sqrt{x} - 1} ight) = \cos^{-1}\left( \frac{\sqrt{x} - 1}{\sqrt{x} + 1} ight)$. Substituting this into the original equation: $y = \cos^{-1}\left( \frac{\sqrt{x} - 1}{\sqrt{x} + 1} ight) + \sin^{-1}\left( \frac{\sqrt{x} - 1}{\sqrt{x} + 1} ight)$ Using the identity $\sin^{-1}(	heta) + \cos^{-1}(	heta) = \frac{\pi}{2}$,we get: $y = \frac{\pi}{2}$ Since $\frac{\pi}{2}$ is a constant,its derivative with respect to $x$ is: $\frac{dy}{dx} = 0$

If $y = \sec^{-1}\left( \frac{\sqrt{x} + 1}{\sqrt{x} - 1} \right) + \sin^{-1}\left( \frac{\sqrt{x} - 1}{\sqrt{x} + 1} \right)$,then $\frac{dy}{dx} = $

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