Answer
$-\dfrac{2\sqrt3}{3}$
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:
$\sec{(\frac{5\pi}{6})}=-\dfrac{2}{\sqrt3} \cdot \dfrac{\sqrt3}{\sqrt3}=-\dfrac{2\sqrt3}{3}$
