Answer
See below
Work Step by Step
Substitute $(-6,9)$ into $x^2+y^2=r^2$ to find $r$:
$$x^2+y^2=r^2\\9^2+(-6)^2=r^2\\r^2=117\\r=\pm 3\sqrt 13\\r=3\sqrt 13$$
$\sin \theta=\frac{y}{r}=\frac{9}{3\sqrt 13}=\frac{3\sqrt 13}{13}$
$\cos \theta=\frac{x}{r}=\frac{-2\sqrt 13}{13}$
$\csc \theta=\frac{r}{y}=\frac{-2\sqrt 2}{2}=\frac{\sqrt 13}{3}$
$\sec \theta=\frac{r}{x}=-\frac{\sqrt 13}{2}$
$\tan \theta=\frac{y}{x}=-\frac{3}{2}$
$\cot \theta=\frac{x}{y}=-\frac{2}{3}$
