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