Question

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

Answer 1

We know ${\sec }^{-1}z={\cos }^{-1}\left(1/z\right)$ and ${\cos }^{-1}y+{\sin }^{-1}y=\pi /2.$

$\therefore y={\cos }^{-1}\frac{\sqrt{x}-1}{\sqrt{\left(x\right)}+1}+{\sin }^{-1}\frac{\sqrt{x}-1}{\sqrt{\left(x\right)}+1}=\frac{\pi }{2}$

$\therefore \frac{dy}{dx}=0.$