Answer
$\sin\theta$ = $ \frac{\sqrt 7}{4}$
$\cos\theta$ = -$ \frac{3}{4}$
$\tan\theta$ = - $ \frac{\sqrt 7}{3}$
$\cot\theta$ = - $ \frac{3}{\sqrt 7}$
$\sec\theta$ = -$ \frac{4}{3}$
$\csc\theta$ =$ \frac{4}{\sqrt 7}$
Work Step by Step
Given, point $(-3, \sqrt 7)$ is on the terminal side of $\theta$, we may apply Definition I to find all six trigonometric functions-
We got $ x = -3, y = \sqrt 7$
Therefore r= $\sqrt (x^{2} + y^{2})$
= $\sqrt ((-3)^{2} + (\sqrt 7)^{2})$
= $\sqrt (9 + 7)$
= $\sqrt (16)$ = 4
i.e. $ x = -3, y = \sqrt 7,$ and $ r= 4$
Applying Definition I-
$\sin\theta$ =$ \frac{y}{r}$ = $ \frac{\sqrt 7}{4}$
$\cos\theta$ =$ \frac{x}{r}$ =$ \frac{-3}{4}$ = -$ \frac{3}{4}$
$\tan\theta$ =$ \frac{y}{x}$ =$ \frac{\sqrt 7}{-3}$ = - $ \frac{\sqrt 7}{3}$
$\cot\theta$ =$ \frac{x}{y}$ =$ \frac{-3}{\sqrt 7}$ = - $ \frac{3}{\sqrt 7}$
$\sec\theta$ =$ \frac{r}{x}$ =$ \frac{4}{-3}$ = -$ \frac{4}{3}$
$\csc\theta$ =$ \frac{r}{y}$ =$ \frac{4}{\sqrt 7}$
