## Precalculus (6th Edition) Blitzer

$\sin \theta =\dfrac{y}{r}=\dfrac{-4}{5} \\ \cos \theta =\dfrac{x}{r}=\dfrac{-3}{5} \\ \tan \theta =\dfrac{y}{x}=\dfrac{4}{3}$ and $\csc \theta =\dfrac{r}{y}=\dfrac{-5}{4} \\ \sec \theta =\dfrac{r}{x}=\dfrac{-5}{3} \\ \cot \theta =\dfrac{x}{y}=\dfrac{3}{4}$
Here, $x= -3; r=5$ $r=\sqrt {x^2+y^2}$ This gives: $y=-\sqrt {(5)^2-(-3)^2}=-4$; Because $\theta$ lies in Quadrant-III. The trigonometric ratios are as follows: $\sin \theta =\dfrac{y}{r}=\dfrac{-4}{5} \\ \cos \theta =\dfrac{x}{r}=\dfrac{-3}{5} \\ \tan \theta =\dfrac{y}{x}=\dfrac{4}{3}$ and $\csc \theta =\dfrac{r}{y}=\dfrac{-5}{4} \\ \sec \theta =\dfrac{r}{x}=\dfrac{-5}{3} \\ \cot \theta =\dfrac{x}{y}=\dfrac{3}{4}$