Answer
$2$
Work Step by Step
The angle $-\dfrac{11\pi}{6}$ is in Quadrant I.
Its reference angle is $\dfrac{\pi}{6}$.
Note that $\dfrac{\pi}{6}$ is a special angle and that $\sin{(\frac{\pi}{6})}=0.5$.
Recall that $\csc{\theta} = \dfrac{1}{\sin{\theta}}$.
Thus, $\csc{\frac{\pi}{6}}=\frac{1}{0.5}=2$.
The trigonometric function values of an angle and its reference angle are either the same or differ only in signs.
Since $-\dfrac{11\pi}{6}$ terminates in Quadrant I, then its cosecant is positive.
Therefore,
$\csc{(-\frac{11\pi}{6})}= 2$.
