Answer
$120^{\circ}$ and $300^{\circ}$
Work Step by Step
$\cot\theta=-\frac{\sqrt 3}{3}$
$\tan\theta=\frac{1}{\cot\theta}=-\frac{3}{\sqrt 3}=-\sqrt 3$
The value of $\tan \theta$ is negative, so $\theta$ may lie in either quadrant II or IV.
We know that $\tan 60^{\circ}=\sqrt 3$.
Recall: $-\tan x= \tan(180^{\circ}-x)$ and $-\tan x= \tan(360^{\circ}-x)$
$\implies -\tan 60^{\circ}=-\sqrt {3}=\tan(180^{\circ}-60^{\circ})$
$=\tan 120^{\circ}$
$120^{\circ}$ which is in the second quadrant is one value of $\theta$.
Now $-\tan 60^{\circ}=-\sqrt 3= \tan (360^{\circ}-60^{\circ})$
$=\tan 300^{\circ}$
$300^{\circ}$ is in the fourth quadrant.
The values of $\theta$ are $ 120^{\circ}$ and $300^{\circ}$.
