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 $\frac{dy}{dx}=$

Answer 1

Given $y={\text{sec}}^{-1}(\frac{\sqrt{x}+1}{\sqrt{x}-1})+{\text{sin}}^{-1}(\frac{\sqrt{x}-1}{\sqrt{x}+1})={\text{cos}}^{-1}(\frac{\sqrt{x}-1}{\sqrt{x}+1})+{\text{sin}}^{-1}(\frac{\sqrt{x}-1}{\sqrt{x}+1})=\frac{\pi }{2}$

$⟹\frac{dy}{dx}=0$