Answer
$\sin\theta$ = $ \frac{-1}{\sqrt {10}}$
$\cos\theta$ =$ \frac{-3}{\sqrt {10}}$
$\tan\theta$ = $ \frac{1}{3}$
$\cot\theta$ = $ 3$
$\sec\theta$ = -$ \frac{\sqrt {10}}{3}$
$\csc\theta$ = -$ \sqrt {10}$
Work Step by Step
Given, point (-3, -1) is on the terminal side of $\theta$ in standard position, we may apply Definition I to find all six trigonometric functions-
We got $ x = -3, y = -1$
Therefore r= $\sqrt {x^{2} + y^{2}}$
= $\sqrt {(-3)^{2} + (-1)^{2}}$
= $\sqrt {9 + 1}$
= $\sqrt {10}$
i.e. $ x = -3$, $y = -1,$ and $ r= \sqrt {10}$
Applying Definition I-
$\sin\theta$ =$ \frac{y}{r}$ = $ \frac{-1}{\sqrt {10}}$
$\cos\theta$ =$ \frac{x}{r}$ =$ \frac{-3}{\sqrt {10}}$
$\tan\theta$ =$ \frac{y}{x}$ =$ \frac{-1}{-3}$ = $ \frac{1}{3}$
$\cot\theta$ =$ \frac{x}{y}$ =$ \frac{-3}{-1}$ = $ 3$
$\sec\theta$ =$ \frac{r}{x}$ =$ \frac{\sqrt {10}}{-3}$ = -$ \frac{\sqrt {10}}{3}$
$\csc\theta$ =$ \frac{r}{y}$ =$ \frac{\sqrt {10}}{-1}$= -$ \sqrt {10}$
