Answer
See below
Work Step by Step
Substitute $(-3,-5)$ into $x^2+y^2=r^2$ to find $r$:
$$x^2+y^2=r^2\\(-3)^2+(-5)^2=r^2\\r^2=34\\r=\pm \sqrt 34\\r=\sqrt 34$$
$\sin \theta=\frac{y}{r}=\frac{-5}{\sqrt 34}=\frac{-5\sqrt 34}{34}$
$\cos \theta=\frac{x}{r}=\frac{-2\sqrt 13}{13}$
$\csc \theta=\frac{r}{y}=\frac{\sqrt 34}{-5}=\frac{-\sqrt 34}{5}$
$\sec \theta=\frac{r}{x}=-\frac{\sqrt 34}{3}$
$\tan \theta=\frac{y}{x}=\frac{5}{3}$
$\cot \theta=\frac{x}{y}=\frac{3}{5}$
