Answer
$sin\theta= \frac{\sqrt {10}}{10}$,
$cos\theta= -\frac{3\sqrt {10}}{10}$,
$cot\theta= -3$,
$sec\theta= -\frac{\sqrt {10}}{3}$,
$csc\theta= \sqrt {10} $.
Work Step by Step
Given $tan\theta=-\frac{1}{3}$, let $x=3,y=1$, we can get $r=\sqrt {x^2+y^2}=\sqrt {10}$. Use the fact that $sin\theta\gt0,tan\theta\lt0\longrightarrow \theta$ is in quadrant II to determine the sign of each function:
$sin\theta=\frac{y}{r}=\frac{\sqrt {10}}{10}$,
$cos\theta=-\frac{x}{r}=-\frac{3\sqrt {10}}{10}$,
$cot\theta=-\frac{x}{y}=-3$,
$sec\theta=-\frac{r}{x}=-\frac{\sqrt {10}}{3}$,
$csc\theta=\frac{r}{y}=\sqrt {10} $.
