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 (-9a, -12a) is on the terminal side of $\theta$, Given that 'a' is a positive number, 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$
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{1}{tan\theta}$ = $ \frac{3}{4}$
$\sec\theta$ =$ \frac{1}{cos\theta}$ = -$ \frac{5}{3}$
$cosec\theta$ =$ \frac{1}{sin\theta}$ = -$ \frac{5}{4}$
