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 $sin\theta=-\frac{5}{13}$, let $y=5, r=13$. We have $x=\sqrt {r^2-y^2}=12$. As $ \theta $ is in quadrant IV, the signs of the other functions are also known. We have:
$sin\theta=-\frac{y}{r}=-\frac{5}{13}$,
$cos\theta=\frac{x}{r}=\frac{12}{13}$,
$tan\theta=-\frac{y}{x}=-\frac{5}{12}$,
$cot\theta=-\frac{x}{y}=-\frac{12}{5}$,
$sec\theta=\frac{r}{x}=\frac{13}{12}$,
$csc\theta=-\frac{r}{y}=-\frac{13}{5}$.
