Solve for real $x$ : ${\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}=\frac{\pi }{4}$. Mention any restrictions on $x$.

Step-by-step solution

1. Let $\theta ={\tan }^{-1}x$, so $x=\tan \theta$ and $\theta \in (-\frac{\pi }{2},\frac{\pi }{2})$.
2. Then ${\tan }^{-1}\frac{1}{x}={\tan }^{-1}(\cot \theta )$. For $\theta >0$, $\cot \theta =\tan \left(\frac{\pi }{2}-\theta \right)$ and the principal value is $\frac{\pi }{2}-\theta$.
3. Hence, for $x>0$ (i.e., $\theta >0$), ${\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}=\theta +\left(\frac{\pi }{2}-\theta \right)=\frac{\pi }{2}$.
4. For $x<0$ (i.e., $\theta <0$), $\cot \theta$ is negative and its principal value is $-\frac{\pi }{2}-\theta$. Thus the sum becomes $\theta +\left(-\frac{\pi }{2}-\theta \right)=-\frac{\pi }{2}$.
5. In both cases the left‑hand side equals $\pm \frac{\pi }{2}$, never $\frac{\pi }{4}$.
6. Since $x=0$ makes $\frac{1}{x}$ undefined, it is excluded.
7. Therefore the given equation has no real solution.

Concept tested: Sum of inverse tangents and principal values

Published 4 September 2026 · How this solution was prepared and checked