Find the principal value of the equation $\tan \theta = - \dfrac{1}{\sqrt{3}}$
$\tan \theta = - \dfrac{1}{\sqrt{3}} < 0$
$\therefore \;$ $\theta$ lies in the second or fourth quadrants.
$\therefore \;$ $\theta = \pi - \dfrac{\pi}{6} = \dfrac{5 \pi}{6}$ $\;$ or $\;$ $\theta = 2 \pi - \dfrac{\pi}{6} = \dfrac{11 \pi}{6}$
Principal value of $\tan \theta$ is in $\left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ $\;$ i.e. $\;$ in first or fourth quadrants.
$\because \;$ $\dfrac{5 \pi}{6} \notin \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$,
$\therefore \;$ Principal value of the given equation is $\dfrac{11 \pi}{6}$ $\;$ i.e. $\;$ $- \dfrac{\pi}{6}$