Question

Find the derivative of ${\sec }^{-1}\left(\frac{x+1}{x-1}\right)+{\sin }^{-1}\left(\frac{x-1}{x+1}\right)$

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

Answer 1

The correct option is A $0$

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