# How do you find the five remaining trigonometric function satisfying csctheta=-3/2, costheta<0?

$\csc t = \frac{1}{\sin t} = - \frac{3}{2}$ --> $\sin t = - \frac{2}{3}$
$\sin t = - \frac{2}{3}$
${\cos}^{2} t = 1 - {\sin}^{2} t = 1 - \frac{4}{9} = \frac{5}{9}$ --> $\cos t = \pm \frac{\sqrt{5}}{3}$
$\cos t = - \frac{\sqrt{5}}{3}$ (cos t < 0)
$\tan t = \frac{\sin}{\cos} = \left(- \frac{2}{3}\right) \left(- \frac{3}{\sqrt{5}}\right) = \frac{2}{\sqrt{5}} = \frac{2 \sqrt{5}}{5}$
$\cot t = \frac{1}{\tan t} = \frac{5}{2 \sqrt{5}}$.
$\sec t = \frac{1}{\cos} = - \frac{3}{\sqrt{5}} = - \frac{3 \sqrt{5}}{5}$.
$\csc t = - \frac{3}{2}$