## Precalculus (6th Edition) Blitzer

$sin\theta=\frac{y}{r}=-\frac{5\sqrt {29}}{29}$ $cos\theta=\frac{x}{r}=-\frac{2\sqrt {29}}{29}$ $tan\theta=\frac{y}{x}=\frac{5}{2}$ $cot\theta=\frac{x}{y}=\frac{2}{5}$ $sec\theta=\frac{r}{x}=-\frac{\sqrt {29}}{2}$ $csc\theta=\frac{r}{y}=-\frac{\sqrt {29}}{5}$
Step 1. Given a point on the terminal side of angle as $(-2,-5)$, we have $x=-2, y=-5$, and $r=\sqrt {(-2)^2+(-5)^2}=\sqrt {29}$ Step 2. By definition, we have $sin\theta=\frac{y}{r}=-\frac{5\sqrt {29}}{29}$ Step 3. By definition, we have $cos\theta=\frac{x}{r}=-\frac{2\sqrt {29}}{29}$ Step 4. By definition, we have $tan\theta=\frac{y}{x}=\frac{5}{2}$ Step 5. By definition, we have $cot\theta=\frac{x}{y}=\frac{2}{5}$ Step 6. By definition, we have $sec\theta=\frac{r}{x}=-\frac{\sqrt {29}}{2}$ Step 7. By definition, we have $csc\theta=\frac{r}{y}=-\frac{\sqrt {29}}{5}$