Answer
$2$
Work Step by Step
Suppose $\theta =\cos^{-1} (-\dfrac{\sqrt 3}{2})$
This gives: $\cos \theta=-\dfrac{\sqrt 3}{2}$
Since, $ r=\sqrt {x^2+y^2}$
This implies that $ y=\sqrt {r^2-x^2}=\sqrt {2^2-(\sqrt 3)^2}= 1$
Therefore, we have $\csc \theta =\csc [\cos^{-1} (-\dfrac{\sqrt 3}{2})]=2$
