Answer
$\sin\theta$ = - $ \frac{4}{5}$
$\cos\theta$ = -$ \frac{3}{5}$
$\tan\theta$ = $ \frac{4}{3}$
$\cot\theta$ = $ \frac{3}{4}$
$\sec\theta$ = -$ \frac{5}{3}$
$\csc\theta$ = -$ \frac{5}{4}$
Work Step by Step
Given, point (-3, -4) is on the terminal side of $\theta$, we may apply Definition I to find all six trigonometric functions-
We got $ x = -3, y = -4$
Therefore r= $\sqrt (x^{2} + y^{2})$
= $\sqrt ((-3)^{2} + (-4)^{2})$
= $\sqrt (9 + 16)$
= $\sqrt (25)$ = 5
i.e. $ x = -3, y = -4,$ and $ r= 5$
Applying Definition I-
$\sin\theta$ =$ \frac{y}{r}$ = $ \frac{-4}{5}$ = - $ \frac{4}{5}$
$\cos\theta$ =$ \frac{x}{r}$ =$ \frac{-3}{5}$ = -$ \frac{3}{5}$
$\tan\theta$ =$ \frac{y}{x}$ =$ \frac{-4}{-3}$ = $ \frac{4}{3}$
$\cot\theta$ =$ \frac{x}{y}$ =$ \frac{-3}{-4}$ = $ \frac{3}{4}$
$\sec\theta$ =$ \frac{r}{x}$ =$ \frac{5}{-3}$ = -$ \frac{5}{3}$
$\csc\theta$ =$ \frac{r}{y}$ =$ \frac{5}{-4}$= -$ \frac{5}{4}$
