Find the value of $\tan^{-1} \left( -\frac{1}{\sqrt{3}} \right) + \cot^{-1} \left( \frac{1}{\sqrt{3}} \right) + \tan^{-1} \left[ \sin \left( -\frac{\pi}{2} \right) \right]$.

$-\frac{\pi}{12}$

The correct answer is Option (3) → $-\frac{\pi}{12}$ ##

We have, $\tan^{-1} \left( -\frac{1}{\sqrt{3}} \right) + \cot^{-1} \left( \frac{1}{\sqrt{3}} \right) + \tan^{-1} \left[ \sin \left( -\frac{\pi}{2} \right) \right]$

$= \tan^{-1} \left( -\tan \frac{\pi}{6} \right) + \cot^{-1} \left( \cot \frac{\pi}{3} \right) + \tan^{-1} \left[ -\sin \frac{\pi}{2} \right]$

$= \tan^{-1} \left[ \tan \left( -\frac{\pi}{6} \right) \right] + \frac{\pi}{3} + \tan^{-1} (-1)$

$= \frac{-\pi}{6} + \frac{\pi}{3} + \tan^{-1} \left( -\tan \frac{\pi}{4} \right) = \frac{-\pi}{6} + \frac{\pi}{3} + \tan^{-1} \left[ \tan \left( -\frac{\pi}{4} \right) \right]$

$= -\frac{\pi}{6} + \frac{\pi}{3} - \frac{\pi}{4} = \frac{-2\pi + 4\pi - 3\pi}{12} = \frac{-5\pi + 4\pi}{12} = -\frac{\pi}{12}$