Answer
$\sin\theta$ = -$ \frac{1}{2}$
$\cos\theta$ =$ \frac{\sqrt 3}{2}$
$\tan\theta$ = - $ \frac{1}{\sqrt 3}$
$\cot\theta$ = - $ \sqrt 3$
$\sec\theta$ =$ \frac{2}{\sqrt 3}$
$\csc\theta$ = - 2
Work Step by Step
Given, point $(\sqrt 3, -1)$ is on the terminal side of $\theta$, we may apply Definition I to find all six trigonometric functions-
We got $ x = \sqrt 3, y = -1$
Therefore r= $\sqrt (x^{2} + y^{2})$
= $\sqrt ((\sqrt 3)^{2} + (-1)^{2})$
= $\sqrt (3 + 1)$
= $\sqrt (4)$ = 2
i.e. $ x = \sqrt 3, y = -1,$ and $ r= 2$
Applying Definition I-
$\sin\theta$ =$ \frac{y}{r}$ = $ \frac{-1}{2}$ = -$ \frac{1}{2}$
$\cos\theta$ =$ \frac{x}{r}$ =$ \frac{\sqrt 3}{2}$
$\tan\theta$ =$ \frac{y}{x}$ =$ \frac{-1}{\sqrt 3}$ = - $ \frac{1}{\sqrt 3}$
$\cot\theta$ =$ \frac{x}{y}$ =$ \frac{\sqrt 3}{-1}$ = - $ \sqrt 3$
$\sec\theta$ =$ \frac{r}{x}$ =$ \frac{2}{\sqrt 3}$
$\csc\theta$ =$ \frac{r}{y}$ =$ \frac{2}{-1}$ = - 2
