# How do you find all six trigonometric function of theta if the point (12, -5) is on the terminal side of theta?

Jun 7, 2017

$\tan t = \frac{y}{x} = - \frac{5}{12}$
${\cos}^{2} t = \frac{1}{1 + {\tan}^{2} t} = \frac{1}{1 + \frac{25}{144}} = \frac{144}{169}$
$\cos t = \frac{12}{13}$ (t is in Quadrant 4)
$\sin t = \tan t . \cos t = \left(- \frac{5}{12}\right) \left(\frac{12}{13}\right) = - \frac{5}{13}$
$\cot t = \frac{1}{\tan} = - \frac{12}{5}$
$\sec t = \frac{1}{\cos} = \frac{13}{12}$
$\csc t = \frac{1}{\sin} = - \frac{13}{5}$