If $y=\operatorname{cosec}^1\left[\frac{\sqrt{x}+1}{\sqrt{x-1}}\right]+\cos…
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 $\frac{d y}{d x}=$
- 0
- 1
- $\frac{2}{\sqrt{x}+1}$
- $\frac{1}{2(\sqrt{\mathrm{x}}+1)}$
Solution
$\begin{aligned}
& y=\operatorname{cosec}^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)+\cos ^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)=\sin ^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)+\cos ^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)=\frac{\pi}{2} \\
& \therefore \frac{d y}{d x}=0
\end{aligned}$
Asked in: MHT CET 2021 (21 Sep Shift 1)
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