If $y=\sec ^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)+\sin ^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$, then $\frac{d y}{d x}$ is :
Answer & explanation
Correct answer: option 4
$y =\sec ^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)+\sin ^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$
$=\cos ^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)+\sin ^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)=\frac{\pi}{2}$
$\Rightarrow \frac{d y}{d x}=0$
Hence (4) is correct answer.