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

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(A) Given $y = \sec^{-1}\left( \frac{x + 1}{x - 1} \right) + \sin^{-1}\left( \frac{x - 1}{x + 1} \right)$.We know that $\sec^{-1}(z) = \cos^{-1}\left(

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

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

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