Answer
$\sin\theta$ = - $ \frac{2}{\sqrt 5}$
$\cos\theta$ = - $ \frac{1}{\sqrt 5}$
$\tan\theta$ = 2
$\cot\theta$ = $ \frac{1}{2}$
$\sec\theta$ = -$ \sqrt 5$
$\csc\theta$ = -$ \frac{\sqrt 5}{2}$
Work Step by Step
Given, point (-1, -2) is on the terminal side of $\theta$, we may apply Definition I to find all six trigonometric functions-
We got $ x = -1, y = -2$
Therefore r= $\sqrt (x^{2} + y^{2})$
= $\sqrt ((-1)^{2} + (-2)^{2})$
= $\sqrt (1 + 4)$
= $\sqrt 5$
i.e. $ x = -1, y = -2,$ and $ r= = \sqrt 5$
Applying Definition I-
$\sin\theta$ =$ \frac{y}{r}$ = $ \frac{-2}{\sqrt 5}$ = - $ \frac{2}{\sqrt 5}$
$\cos\theta$ =$ \frac{x}{r}$ =$ \frac{-1}{\sqrt 5}$ = - $ \frac{1}{\sqrt 5}$
$\tan\theta$ =$ \frac{y}{x}$ =$ \frac{-2}{-1}$ = 2
$\cot\theta$ =$ \frac{x}{y}$ =$ \frac{-1}{-2}$ = $ \frac{1}{2}$
$\sec\theta$ =$ \frac{r}{x}$ =$ \frac{\sqrt 5}{-1}$ = -$ \sqrt 5$
$\csc\theta$ =$ \frac{r}{y}$ =$ \frac{\sqrt 5}{-2}$ = -$ \frac{\sqrt 5}{2}$
