## Trigonometry 7th Edition

$2$
$\dfrac{5\pi}{6}$ 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 $\csc{\theta}$ is the reciprocal of the sine function, we first have to find the value of $\sin{\theta}$. $\dfrac{\pi}{6}$ is a special angle whose sine value is $\dfrac{1}{2}$. Recall that an angle and its reference angle have the same sine values, except possibly in their signs. Since $\dfrac{5\pi}{6}$ is in Quadrant II, it sine value is positive. Thus, $\sin{(\frac{5\pi}{6})}=\frac{1}{2}$. RECALL: $\csc{\theta} = \dfrac{1}{\sin{\theta}}$ Therefore, $\csc{(\frac{5\pi}{6})} = \dfrac{1}{\sin{(\frac{5\pi}{6}})}=\dfrac{1}{\frac{1}{2}} = 1 \cdot \frac{2}{1} = 2$