Question

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

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

Answer 1

Answer: (A)

$0$

$y={\sec }^{-1}(\frac{x+1}{x-1})+{\sin }^{-1}(\frac{x-1}{x+1})$

We know that

${\sec }^{-1}\left(t\right)={\cos }^{-1}\left(\frac{1}{t}\right)$

and ${\cos }^{-1}\left(t\right)+{\sin }^{-1}\left(t\right)=\frac{\pi }{2}$

$\therefore y={\cos }^{-1}(\frac{x-1}{x+1})+{\sin }^{-1}(\frac{x-1}{x+1})$

$=\frac{\pi }{2}$

$\therefore \frac{dy}{dx}=\frac{d}{dx}\left(\frac{\pi }{2}\right)$

$=0$