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If $y=\operatorname{cosec}^1\left[\frac{\sqrt{x}+1}{\sqrt{x-1}}\right]+\cos ^{-1}\left[\frac{\sqrt{x}-1}{\sqrt{x}+1}\right]$, then $\frac{d y}{d x}=$
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$$
\begin{aligned}
& y=\operatorname{cosec}^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)+\cos ^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)=\sin ^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)+\cos ^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)=\frac{\pi}{2} \\
& \therefore \frac{d y}{d x}=0
\end{aligned}
$$
\begin{aligned}
& y=\operatorname{cosec}^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)+\cos ^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)=\sin ^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)+\cos ^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)=\frac{\pi}{2} \\
& \therefore \frac{d y}{d x}=0
\end{aligned}
$$
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