Question

If x < 0, then tan^-1x + tan^-1$\frac{1}{x}$ is equal to ____________________.

Answer 1

We know

${\tan }^{-1}\frac{1}{x}=\left\{\begin{array}{ll}{\cot }^{-1}x,& \mathrm{for}x>0\\ -\mathrm{\pi }+{\cot }^{-1}x,& \mathrm{for}x<0\end{array}\right.$

$$\begin{gathered} \therefore {\tan }^{-1}x+{\tan }^{-1}\frac{1}{x} \\ ={\tan }^{-1}x+{\cot }^{-1}x-\mathrm{\pi }\left(x<0\right) \\ =\frac{\mathrm{\pi }}{2}-\mathrm{\pi }\left({\tan }^{-1}\mathrm{x}+{\cot }^{-1}\mathrm{x}=\frac{\mathrm{\pi }}{2},\forall x\in \mathbf{R}\right) \\ =-\frac{\mathrm{\pi }}{2} \end{gathered}$$

If x < 0, then tan⁻¹x + tan⁻¹$\frac{1}{x}$ is equal to $\underline{-\frac{\mathrm{\pi }}{2}}$.