Question

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

Answer 1

Given $y=se{c}^{-1}\left(\frac{x+1}{x-1}\right)+si{n}^{-1}\left(\frac{x-1}{x+1}\right)$

Let $se{c}^{-1}\left(\frac{x+1}{x-1}\right)=t$

$sect=\frac{x+1}{x-1}$

$cost=\frac{x-1}{x+1}$

$\therefore t=co{s}^{-1}\left(\frac{x-1}{x+1}\right)$

$\therefore y=co{s}^{-1}\left(\frac{x-1}{x+1}\right)+si{n}^{-1}\left(\frac{x-1}{x+1}\right)$

$y=\frac{\pi }{2}$ $(\because co{s}^{-1}\theta +si{n}^{-1}\theta =\frac{\pi }{2})$

$(\forall \theta \in [-1,1\left]\right)$

$\therefore \frac{dy}{dx}=0$ $\because \frac{x-1}{x+1}$ is lies between [-1,1]