Answer
$\text{Quadrant II}$
Work Step by Step
$P= (x,y) \text{ is the point on the unit circle corresponding to } \theta$.
$\sec{\theta} = \dfrac{1}{x}<0 \hspace{20pt} \therefore x<0$
We have:
$\sin{\theta} = y >0$
Note that points in Quadrant II have $y\gt 0 \text{ and } x\lt 0$.
Thus, $\theta \in \text{Quadrant II}$.
