# The point (3 1/2, -7 3/4) is on the terminal side of an angle in standard position, how do you determine the exact values of the six trigonometric functions of the angle?

The point $\left(x = \frac{7}{2} , y = - \frac{31}{4}\right)$, on the terminal side of the angle (or arc) t, lies in Quadrant 4.
$\tan t = \frac{y}{x} = \frac{7}{2} : - \frac{31}{4} = \left(\frac{7}{2}\right) \left(- \frac{4}{31}\right) = - \frac{14}{31}$
${\cos}^{2} t = \frac{1}{1 + {\tan}^{2} t} = \frac{1}{1 + \frac{196}{961}} = \frac{961}{1157}$
$\cos t = \frac{31}{\sqrt{1157}}$ (because t lies in Quadrant 4)
$\sin t = \cos t . \tan t = \left(\frac{31}{\sqrt{1157}}\right) \left(- \frac{14}{31}\right) = - \frac{14}{\sqrt{1157}}$
$\cot t = \frac{1}{\tan} = - \frac{31}{14}$
$\sec t = \frac{1}{\cos} = \frac{\sqrt{1157}}{31}$
$\csc t = \frac{1}{\sin} = - \frac{\sqrt{1157}}{14}$