## Precalculus (6th Edition) Blitzer

Here, $x= -4; y=3$ $r=\sqrt {x^2+y^2}=\sqrt {(-4)^2+(3)^2}=5$ The trigonometric ratios are as follows: $sin \theta =\dfrac{y}{r}=\dfrac{3}{5} \\cos \theta =\dfrac{x}{r}=\dfrac{-4}{5} \\ \tan \theta =\dfrac{y}{x}=\dfrac{-3}{4}$ and $\csc \theta =\dfrac{r}{y}=\dfrac{5}{3} \\ \sec \theta =\dfrac{r}{x}=\dfrac{-5}{4} \\ \cot \theta =\dfrac{x}{y}=\dfrac{-4}{3}$
Here, $x= -12; y=5$ $r=\sqrt {x^2+y^2}=\sqrt {(-12)^2+(5)^2}=13$ The trigonometric ratios are as follows: $sin \theta =\dfrac{y}{r}=\dfrac{5}{13} \\cos \theta =\dfrac{x}{r}=\dfrac{-12}{13} \\ \tan \theta =\dfrac{y}{x}=\dfrac{-5}{12}$ and $\csc \theta =\dfrac{r}{y}=\dfrac{13}{5} \\ \sec \theta =\dfrac{r}{x}=\dfrac{-13}{12} \\ \cot \theta =\dfrac{x}{y}=\dfrac{-12}{5}$