Answer
$\sin\theta=\frac{y}{r}=\frac{-\sqrt 3}{2}$
$\cos\theta=\frac{x}{r}=\frac{-1}{2}$
$\tan\theta=\frac{y}{x}=\frac{-\sqrt 3 }{-1}=\sqrt 3$
$\cot\theta=\frac{x}{y}=\frac{-1}{-\sqrt 3}=\frac{\sqrt 3}{3}$
$\sec\theta=\frac{r}{x}=\frac{2}{-1}=-2$
$\csc\theta=\frac{r}{y}=\frac{2}{-\sqrt 3}=\frac{-2\sqrt 3}{3}$
Work Step by Step
1. Angle in III quadrant, therefore x and y are neagtive
$x=-1$ and $y=-\sqrt 3$
2. Calculate r using distance formula:
$r=\sqrt {(-1)^{2}+(-\sqrt 3)^{2}} =\sqrt {4}=2$
3. Plug the values into the expression
$\sin\theta=\frac{y}{r}=\frac{-\sqrt 3}{2}$
$\cos\theta=\frac{x}{r}=\frac{-1}{2}$
$\tan\theta=\frac{y}{x}=\frac{-\sqrt 3 }{-1}=\sqrt 3$
$\cot\theta=\frac{x}{y}=\frac{-1}{-\sqrt 3}=\frac{\sqrt 3}{3}$
$\sec\theta=\frac{r}{x}=\frac{2}{-1}=-2$
$\csc\theta=\frac{r}{y}=\frac{2}{-\sqrt 3}=\frac{-2\sqrt 3}{3}$
