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

$\text{Quadrant II}$
$P= (x,y) \text{ is the point on the unit circle corresponding to } \theta$. 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, $\theta \in \text{Quadrant II}$ .