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 $tan\theta=\frac{1}{3}$ and $\theta$ is in quadrant III, let $x=-3, y=-1$, we have $r=\sqrt {(-3)^2+(-1)^2}=\sqrt {10}$, thus:
$sin\theta=\frac{y}{r}=-\frac{\sqrt {10}}{10}$,
$cos\theta=\frac{x}{r}=-\frac{3\sqrt {10}}{10}$,
$cot\theta=\frac{1}{tan\theta}=3$,
$sec\theta=\frac{1}{cos\theta}=-\frac{\sqrt {10}}{3}$.
$csc\theta=\frac{1}{sin\theta}=-\sqrt {10}$.
