Question

If $y=se{c}^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)+{\sin }^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$, then $\frac{dy}{dx}$ equals

• A. $1$
• B. $0$
• C. $\frac{\sqrt{x}+1}{\sqrt{x}-1}$
• D. $\frac{\sqrt{x}-1}{\sqrt{x}+1}$

Answer 1

The correct option is B $0$

$y=se{c}^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)+{\sin }^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$
$y={\cos }^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)+{\sin }^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)=\frac{\pi }{2}$ ............. $\because {\sin }^{-1}\theta +{\cos }^{-1}\theta =\frac{\pi }{2}$
$\therefore \frac{dy}{dx}=\frac{d}{dx}\left(\frac{\pi }{2}\right)=0$