Chapter 1 - Trigonometric Functions - Section 1.4 Using the Definitions of the Trigonometric Functions - 1.4 Exercises - Page 38: 70

$\sqrt {3}$

Recall the identity $1+\cot^{2}\theta=\csc^{2}\theta$ Knowing that $\csc\theta=-2$, we get $1+\cot^{2}\theta=(-2)^{2}=4$ $\implies \cot^{2}\theta=4-1=3$ Or $\cot\theta=\pm \sqrt {3}$ In the third quadrant, $\cot\theta$ is positive. Therefore, $\cot\theta=\sqrt {3}$