Write the value of $\tan^{-1}(\sqrt{3}) + \cot^{-1}(-\sqrt{3})$.

$\frac{7\pi}{6}$

The correct answer is Option (2) → $\frac{7\pi}{6}$ ##

$\tan^{-1} \sqrt{3} + \cot^{-1}(-\sqrt{3})$

$ = \tan^{-1} \left( \tan \frac{\pi}{3} \right) + (\pi - \cot^{-1} \sqrt{3})$

$= \frac{\pi}{3} + \pi - \cot^{-1} \sqrt{3} = \frac{4\pi}{3} - \frac{\pi}{6} = \frac{7\pi}{6}$