## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

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