Question

If $y={\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right),\mathrm{find}\frac{dy}{dx}$.

Answer 1

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

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