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