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If $y=\sec ^{-1}\left(\frac{x+1}{x-1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right)$, then $\frac{d y}{d x}$ is equal to
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$\mathrm{y}=\sec ^{-1}\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)+\sin ^{-1}\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)$
$=\cos ^{-1}\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)+\sin ^{-1}\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)$
$=\frac{\pi}{2}\left(\because \sin ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{x}=\frac{\pi}{2}\right) \quad \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=0$
$=\cos ^{-1}\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)+\sin ^{-1}\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)$
$=\frac{\pi}{2}\left(\because \sin ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{x}=\frac{\pi}{2}\right) \quad \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=0$
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