## Precalculus (6th Edition) Blitzer

$-2$
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: $(2\pi - \theta)$ Thus, the reference angle of $\dfrac{ 22 \pi}{3}-\dfrac{21\pi}{3}=\dfrac{ \pi}{3}$ So, $\sec \dfrac{ \pi}{3}=\dfrac{1}{\cos \dfrac{ \pi}{3}}=2$ Thus, we have $\sec \dfrac{\pi}{3}= -2$ because $\theta$ lies in Quadrant-III.
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