Answer
$sin ~\theta = \frac{\sqrt{3}}{2}$
$cos ~\theta = \frac{-1}{2}$
$tan ~\theta = -\sqrt{3}$
$csc ~\theta = \frac{2}{\sqrt{3}}$
$sec ~\theta = -2$
$cot ~\theta = \frac{-1}{\sqrt{3}}$
Work Step by Step
$\sqrt{3}~x+y = 0, x \leq 0$
$y = -\sqrt{3}~x$
$\frac{y}{x} = -\frac{\sqrt{3}}{1}$
Since $x\leq 0$, we can let $x = -1$ and $y = \sqrt{3}$
Then $r = \sqrt{(-1)^2+(\sqrt{3})^2} = 2$
We can find the values of the six trigonometric functions:
$sin ~\theta = \frac{y}{r} = \frac{\sqrt{3}}{2}$
$cos ~\theta = \frac{x}{r} = \frac{-1}{2}$
$tan ~\theta = \frac{y}{x} = \frac{\sqrt{3}}{-1} = -\sqrt{3}$
$csc ~\theta = \frac{r}{y} = \frac{2}{\sqrt{3}}$
$sec ~\theta = \frac{r}{x} = \frac{2}{-1} = -2$
$cot ~\theta = \frac{x}{y} = \frac{-1}{\sqrt{3}}$
