# Chapter 3 - Section 3.2 - Radians and Degrees - 3.2 Problem Set - Page 133: 64

$-\dfrac{2\sqrt3}{3}$

#### Work Step by Step

$\dfrac{5\pi}{6}$ is in Quadrant II. The reference angle of an angle $\theta$ in Quadrant II can be found using the formula $\pi - \theta$. Thus, the reference angle of the given angle is: $=\pi - \dfrac{5\pi}{6} = \dfrac{6\pi}{6}- \dfrac{5\pi}{6} = \dfrac{\pi}{6}$ Since $\sec{\theta}$ is the reciprocal of the cosine function, we first have to find the value of $\cos{\theta}$. $\dfrac{\pi}{6}$ is a special angle whose cosine value is $\dfrac{\sqrt3}{2}$. Recall that an angle and its reference angle have the same cosine values, except possibly in their signs. Since $\dfrac{5\pi}{6}$ is in Quadrant II, it cosine value is negative. Thus, $\cos{(\frac{5\pi}{6})}=-\frac{\sqrt3}{2}$. RECALL: $\sec{\theta} = \dfrac{1}{\cos{\theta}}$ Therefore, $\sec{(\frac{5\pi}{6})} = \dfrac{1}{\cos{(\frac{5\pi}{6}})}=\dfrac{1}{-\frac{\sqrt3}{2}} = 1 \cdot \left(-\dfrac{2}{\sqrt3}\right) = -\dfrac{2}{\sqrt3}$ Rationalize the denominator to obtain: $\sec{(\frac{5\pi}{6})}=-\dfrac{2}{\sqrt3} \cdot \dfrac{\sqrt3}{\sqrt3}=-\dfrac{2\sqrt3}{3}$

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