# How do you find all six trigonometric function of theta if the point (-9a, -12a) is on the terminal side of theta?

Nov 23, 2017

cos t = - 3/5
sin t = - 4/5
cot t = 3/4
sec t = - 5/3
csct = - 5/4

#### Explanation:

t is in Quadrant 3.
$\tan t = \frac{y}{x} = - 12 \frac{a}{-} 9 a = \frac{4}{3}$
${\cos}^{2} t = \frac{1}{1 + {\tan}^{2} t} = \frac{1}{1 + \frac{16}{9}} = \frac{9}{25}$
$\cos t = \pm \frac{3}{5}$
Since t is in Quadrant 3, take the negative answer.
${\sin}^{2} t = 1 - {\cos}^{2} t = 1 - \frac{9}{25} = \frac{16}{25}$.
$\sin t = \pm \frac{4}{5}$
Take the negative answer, because t is in Q. 3.
$\cot t = \frac{3}{4}$
$\sec t = \frac{1}{\cos t} = - \frac{5}{3}$
$\csc t = \frac{1}{\sin t} = - \frac{5}{4}$