Answer
$\sin\theta$ = -$ \frac{5}{13}$
$\cos\theta$ =$ \frac{12}{13}$
$\tan\theta$ = - $ \frac{5}{12}$
$\cot\theta$ = - $ \frac{12}{5}$
$\sec\theta$ =$ \frac{13}{12}$
$\csc\theta$ = - $ \frac{13}{5}$
Work Step by Step
Given, point (12, -5) is on the terminal side of $\theta$, we may apply Definition I to find all six trigonometric functions-
We got $ x = 12, y = -5$
Therefore r= $\sqrt (x^{2} + y^{2})$
= $\sqrt ((12)^{2} + (-5)^{2})$
= $\sqrt (144 + 25)$
= $\sqrt (169)$ = 13
i.e. $ x = 12, y = -5,$ and $ r= 13$
Applying Definition I-
$\sin\theta$ =$ \frac{y}{r}$ = $ \frac{-5}{13}$ = -$ \frac{5}{13}$
$\cos\theta$ =$ \frac{x}{r}$ =$ \frac{12}{13}$
$\tan\theta$ =$ \frac{y}{x}$ =$ \frac{-5}{12}$ = - $ \frac{5}{12}$
$\cot\theta$ =$ \frac{x}{y}$ =$ \frac{12}{-5}$ = - $ \frac{12}{5}$
$\sec\theta$ =$ \frac{r}{x}$ =$ \frac{13}{12}$
$\csc\theta$ =$ \frac{r}{y}$ =$ \frac{13}{-5}$ = - $ \frac{13}{5}$
