Answer
$\sin\theta=\frac{y}{r}=\frac{-2}{\sqrt 5}=-\frac{2\sqrt 5}{5}$
$\cos\theta=\frac{x}{r}=\frac{-1}{\sqrt 5}=-\frac{\sqrt 5}{5}$
$\csc\theta=\frac{r}{y}=\frac{\sqrt 5}{-2}$
$\sec\theta=\frac{r}{x}=\frac{\sqrt 5}{-1}=-\sqrt 5$
$\tan\theta=\frac{y}{x}=\frac{-2}{-1}=2$
$\cot\theta=\frac{x}{y}=\frac{-1}{-2}=\frac{1}{2}$
Work Step by Step
We know that angle is in the III quadrant, therefore x and y are negative, then using tangent of the angle and knowledge about the signs (of x and y) we will find r.
$x= -1; y=-2$
$r=\sqrt {(x)^{2}+(y)^{2}}= \sqrt {(-1)^{2}+(-2)^{2}}=\sqrt 5=\sqrt 5$
