## Trigonometry 7th Edition

$\sin\theta$ = - $\frac{4}{5}$ $\cos\theta$ = -$\frac{3}{5}$ $\tan\theta$ = $\frac{4}{3}$ $\cot\theta$ = $\frac{3}{4}$ $\sec\theta$ = -$\frac{5}{3}$ $\csc\theta$ = -$\frac{5}{4}$
Given, point (-9a, -12a) is on the terminal side of $\theta$, we may apply Definition I to find all six trigonometric functions- We got $x = -9a, y = -12a$ Therefore r= $\sqrt (x^{2} + y^{2})$ = $\sqrt ((-9a)^{2} + (-12a)^{2})$ = $\sqrt (81a^{2} + 144 a^{2})$ = $\sqrt (225 a^{2})$ = 15a i.e. $x = -9a, y = -12a,$ and $r= 15a$ Given that 'a' is a positive number Applying Definition I- $\sin\theta$ =$\frac{y}{r}$ = $\frac{-12a}{15a}$ = - $\frac{4}{5}$ $\cos\theta$ =$\frac{x}{r}$ =$\frac{-9a}{15a}$ = -$\frac{3}{5}$ $\tan\theta$ =$\frac{y}{x}$ =$\frac{-12a}{-9a}$ = $\frac{4}{3}$ $\cot\theta$ =$\frac{x}{y}$ =$\frac{-9a}{-12a}$ = $\frac{3}{4}$ $\sec\theta$ =$\frac{r}{x}$ =$\frac{15a}{-9a}$ = -$\frac{5}{3}$ $\csc\theta$ =$\frac{r}{y}$ =$\frac{15a}{-12a}$= -$\frac{5}{4}$