Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},y={\sec }^{-1}\left(\frac{x+1}{x-1}\right)+{\sin }^{-1}\left(\frac{x-1}{x+1}\right) \\ \Rightarrow y={\cos }^{-1}\left(\frac{x-1}{x+1}\right)+{\sin }^{-1}\left(\frac{x-1}{x+1}\right)\left[\because {\sec }^{-1}x={\cos }^{-1}\left(\frac{1}{x}\right)\right] \\ \Rightarrow y=\frac{\mathrm{\pi }}{2}\left[\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{\mathrm{\pi }}{2}\right] \end{gathered}$$

$\Rightarrow \frac{dy}{dx}=0$