Question

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

Answer 1

The solution is the set of all $x$ such that $-1 \le x \le 1$.

Answer 2

Let $y = \tan^{-1}x$. The principal range of $\tan^{-1}x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$, so $2y \in (-\pi, \pi)$. Substituting $x = \tan y$ into $\frac{2x}{1+x^2}$ gives $\frac{2\tan y}{1+\tan^2 y} = \sin(2y)$. The equation becomes $\sin^{-1}(\sin(2y)) = 2y$. This is valid if $2y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, which means $y \in [-\frac{\pi}{4}, \frac{\pi}{4}]$. Since $y = \tan^{-1}x$, we have $\tan^{-1}x \in [-\frac{\pi}{4}, \frac{\pi}{4}]$. This implies $-1 \le x \le 1$.