## Precalculus (6th Edition) Blitzer

$- \dfrac{2 \sqrt 3}{3}$
The reference angle of an angle $0 \leq \theta \lt 2\pi$ based on its position can be computed by using the following steps: a) Quadrant- I: $\theta$ b) Quadrant -II: $(\pi-\theta)$ c) Quadrant- III: $(\theta - \pi)$ d) Quadrant - IV: $(2\pi - \theta)$ First, we make the angle positive: $-\dfrac{2 \pi}{3}+2\pi=\dfrac{4\pi}{3}$ We are in the 3rd quadrant, so subtract $\pi$: $\dfrac{4\pi}{3}-\pi=\dfrac{\pi}{3}$ Thus, we have $\csc \dfrac{\pi}{3}=- \dfrac{2 \sqrt 3}{3}$ The value is negative because initially we are in the 3rd qauadrant, where $\csc$ is negative.