Answer
$\text{Quadrant II}$
Work Step by Step
$P= (x,y) \text{ is the point on the unit circle corresponding to } \theta$.
Since $\csc{\theta} = \dfrac{1}{y}>0, \text{ then } y>0$
We also know that $\cos{\theta} = x<0$
Note that points in Quadrant II have $y\gt 0 \text{ and } x \lt 0$
Thus, $\theta \in \text{Quadrant II}$
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