Question:
If $x<0$, then $\tan ^{-1} x+\tan ^{-1} \frac{1}{x}$ is equal to ____________________.
Solution:
We know
$\tan ^{-1} \frac{1}{x}= \begin{cases}\cot ^{-1} x, & \text { for } x>0 \\ -\pi+\cot ^{-1} x, & \text { for } x<0\end{cases}$
$\therefore \tan ^{-1} x+\tan ^{-1} \frac{1}{x}$
$=\tan ^{-1} x+\cot ^{-1} x-\pi$ $(x<0)$
$=\frac{\pi}{2}-\pi$ $\left(\tan ^{-1} \mathrm{x}+\cot ^{-1} \mathrm{x}=\frac{\pi}{2}, \forall x \in \mathbf{R}\right)$
$=-\frac{\pi}{2}$
If $x<0$, then $\tan ^{-1} x+\tan ^{-1} \frac{1}{x}$ is equal to $-\frac{\pi}{2}$
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