Question

If $y=sin^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)+sec^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)$, $x > 0$, then $\frac{dy}{dx}$ is equal to

• A. $1$
• B. $0$
• C. $\frac{\pi}{2}$
• D. None of these

Answer 1

Answer: (B)

$y=sin^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)+sec^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)$
$=sin^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$$+cos^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$
$=\frac{\pi}{2}$
$\left\{\because sin^{-1}\,x+cos^{-1}\,x=\frac{\pi}{2}\right\}$
$\Rightarrow y=\frac{\pi}{2}$
$\Rightarrow \frac{dy}{dx}=0$