## Precalculus (6th Edition) Blitzer

$\theta$ lies in the Second Quadrant or Quadrant-II.
The trigonometric ratios are as follows: $\sin \theta =\dfrac{y}{r} \\ \cos \theta =\dfrac{x}{r} \\ \tan \theta =\dfrac{y}{x}\\ \csc \theta =\dfrac{r}{y} \\ \sec \theta =\dfrac{r}{x} \\ \cot \theta =\dfrac{x}{y}$ where, $r=\sqrt {x^2+y^2}$ It has been seen that $x$ is negative and $y$ is positive; this implies that the angle $\theta$ lies in the Second Quadrant or Quadrant-II.