Find the principal value of the equation $\sqrt{3} \cot \theta = 1$
$\sqrt{3} \cot \theta = 1$ $\implies$ $\cot \theta = \dfrac{1}{\sqrt{3}}$ $\implies$ $\tan \theta = \sqrt{3} > 0$
$\therefore \;$ $\theta$ lies in the first or the third quadrants.
Now, $\tan \theta = \sqrt{3} = \tan \left(\dfrac{\pi}{3}\right)$
$\implies$ $\theta = \dfrac{\pi}{3}$, $\;$ $\pi + \dfrac{\pi}{3} = \dfrac{4 \pi}{3}$
Principal value of $\tan \theta$ is in $\left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ $\;$ i.e. $\;$ in first or fourth quadrants.
$\because \;$ $\dfrac{4 \pi}{3} \notin \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$,
$\therefore \;$ Principal value of the given equation is $\dfrac{\pi}{3}$