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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 $\frac{d y}{d x}=$
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$\begin{aligned} & y=\sec ^{-1}\left(\frac{x+1}{x-1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right) \\ & \text { or } y=\cos ^{-1} \frac{x-1}{x+1}+\sin ^{-1}\left(\frac{x-1}{x+1}\right) \\ & \therefore y=\frac{\pi}{2} \Rightarrow \frac{d y}{d x}=0 \quad\left(\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right)\end{aligned}$
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