Solve for real $x$: ${\tan }^{-1}x+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)=\frac{\pi }{2}$. State all possible solutions and any restrictions on $x$.

Step-by-step solution

1. Let $\alpha ={\tan }^{-1}x$, so $x=\tan \alpha$ and $\alpha \in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$.
2. Using the identity $\frac{2\tan \alpha }{1+{\tan }^2\alpha }=\sin 2\alpha$, we have ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)={\sin }^{-1}(\sin 2\alpha )$.
3. Since $2\alpha \in (-\pi ,\pi )$, ${\sin }^{-1}(\sin 2\alpha )=2\alpha$ (principal value).
4. The given equation becomes $\alpha +2\alpha =\frac{\pi }{2}$, i.e. $3\alpha =\frac{\pi }{2}$.
5. Thus $\alpha =\frac{\pi }{6}$, which lies in the allowed range.
6. Now $x=\tan \alpha =\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$.
7. Check the original equation with $x=\frac{1}{\sqrt{3}}$: ${\tan }^{-1}\frac{1}{\sqrt{3}}=\frac{\pi }{6}$ and $\frac{2x}{1+x^2}=\frac{2/\sqrt{3}}{1+1/3}=\frac{2/\sqrt{3}}{4/3}=\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2}$, whose inverse sine is $\frac{\pi }{3}$. Sum $=\frac{\pi }{6}+\frac{\pi }{3}=\frac{\pi }{2}$, so it satisfies the equation.
8. Hence the only real solution is $x=\frac{1}{\sqrt{3}}$, which can be written as $x=\frac{\sqrt{3}}{3}$ or $x=1/\sqrt{3}$.

Concept tested: Solving equations involving inverse trig identities

Published 10 September 2026 · How this solution was prepared and checked