Question:
Find the value of $\tan ^{-1} \sqrt{3}-\cot ^{-1}(-\sqrt{3})$.
Solution:
$\tan ^{-1}(\sqrt{3})-\cot ^{-1}(-\sqrt{3})$
$=\tan ^{-1}\left\{\tan \left(\frac{\pi}{3}\right)\right\}-\cot ^{-1}\left(\cot \frac{5 \pi}{6}\right)$ $\left[\because\right.$ Range of $\tan ^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) ; \frac{\pi}{3} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and range of $\cot ^{-1}$ is $\left.[0, \pi] ; \frac{5 \pi}{6} \in[0, \pi]\right]$
$=\frac{\pi}{3}-\frac{5 \pi}{6}$
$=-\frac{\pi}{2}$