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 $y=1, x=3$. We have $r=\sqrt {x^2+y^2}=\sqrt {10}$. As $ \theta $ is in quadrant III, the signs of the other functions are also known. We have:
$sin\theta=-\frac{y}{r}=-\frac{\sqrt {10}}{10}$,
$cos\theta=-\frac{x}{r}=-\frac{3\sqrt {10}}{10}$,
$tan\theta=\frac{y}{x}=\frac{1}{3}$,
$cot\theta=\frac{x}{y}=3$,
$sec\theta=-\frac{r}{x}=-\frac{\sqrt {10}}{3}$,
$csc\theta=-\frac{r}{y}=-\sqrt {10}$.
