Question

If −1 < x < 0, then write the value of ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$.

Answer 1

Let $x=-\tan y$
where $0<y<\frac{\mathrm{\pi }}{2}$
Then,
$$\begin{gathered} {\sin }^{-1}\left(\frac{2x}{1+x^2}\right)+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)={\sin }^{-1}\left(\frac{-2\tan y}{1+{\tan }^{2}y}\right)+{\cos }^{-1}\left(\frac{1-{\tan }^{2}y}{1+{\tan }^{2}y}\right) \\ ={\sin }^{-1}\left\{-\sin \left(2y\right)\right\}+{\cos }^{-1}\left\{\cos \left(2y\right)\right\} \\ =-{\sin }^{-1}\left\{\sin \left(2y\right)\right\}+{\cos }^{-1}\left\{\cos \left(2y\right)\right\} \\ =-2y+2y \\ =0 \end{gathered}$$

$\therefore {\sin }^{-1}\left(\frac{2x}{1+x^2}\right)+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=0$