# How do you find the values of the six trigonometric functions given sectheta=-2 and sintheta>0?

Feb 6, 2017

$\sec t = \frac{1}{\cos t} = - 2$ --> $\cos t = - \frac{1}{2}$
Find sin t.
${\sin}^{2} t = 1 - {\cos}^{2} t = 1 - \frac{1}{4} = \frac{3}{4}$ --> $\sin t = \pm \frac{\sqrt{3}}{2}$
Since sin t > 0, t is in Quadrant II, there for:
$\sin t = \frac{\sqrt{3}}{2}$.
$\tan t = \frac{\sin}{\cos} = \left(\frac{\sqrt{3}}{2}\right) \left(- \frac{2}{1}\right) = - \sqrt{3}$
$\cot t = \frac{1}{\tan} = - \frac{1}{\sqrt{3}} = - \frac{\sqrt{3}}{3}$
$\csc t = \frac{1}{\sin} = \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}$