Answer
$sin\theta=\frac{12}{13}$,
$cos\theta=-\frac{5}{13}$,
$cot\theta=-\frac{5}{12}$,
$sec\theta=-\frac{13}{5}$,
$csc\theta=\frac{13}{12}$.
Work Step by Step
Given $tan\theta=-\frac{12}{5}$ and $\theta$ in quadrant II, let $x=-5, y=12$, we have $r=\sqrt {(-5)^2+12^2}=13$ and
$sin\theta=\frac{y}{r}=\frac{12}{13}$,
$cos\theta=\frac{x}{r}=-\frac{5}{13}$,
$cot\theta=\frac{1}{tan\theta}=-\frac{5}{12}$,
$sec\theta=\frac{1}{cos\theta}=-\frac{13}{5}$,
$csc\theta=\frac{1}{sin\theta}=\frac{13}{12}$.
