# How do you find the value of tan theta given sec theta=-9/4 and in quadrant III?

Dec 17, 2016

$\frac{\sqrt{65}}{4}$
$\cos t = \frac{1}{\sec t} = - \frac{4}{9}$
${\sin}^{2} t = 1 - {\cos}^{2} t = 1 - {\left(\frac{4}{9}\right)}^{2} = \frac{81 - 16}{81} = \frac{65}{81}$ -->
$\sin t = \pm \frac{\sqrt{65}}{9}$
$\tan t = \sin \frac{t}{\cos t} = \left(- \frac{\sqrt{65}}{9}\right) \left(- \frac{9}{4}\right) = \frac{\sqrt{65}}{4}$