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 $tan\theta=\frac{12}{5}$, let $y=12, x=5$. We have $r=\sqrt {x^2+y^2}=13$. As $sin\theta\lt0, tan\theta\gt0$, we know it is in quadrant III, thus the signs of the other functions are also known. We have:
$sin\theta=-\frac{y}{r}=-\frac{12}{13}$,
$cos\theta=-\frac{x}{r}=-\frac{5}{13}$,
$tan\theta=\frac{y}{x}=\frac{12}{5}$,
$cot\theta=\frac{x}{y}=\frac{5}{12}$,
$sec\theta=-\frac{r}{x}=-\frac{13}{5}$,
$csc\theta=-\frac{r}{y}=-\frac{13}{12}$.
