Question

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

Answer 1

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

We know that

$se{c}^{-1}x=co{s}^{-1}\frac{1}{x}$ [For $x\in (-\infty ,-1]\cup [1,\infty )]$

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

Also $si{n}^{-1}t+co{s}^{-1}t=\frac{\pi }{2}$

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

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