# How do you find the value of sec theta given tan theta=3/2 and in quadrant III?

Aug 8, 2017

$- \frac{\sqrt{13}}{2.}$
${\sec}^{2} \theta = 1 + {\tan}^{2} \theta = 1 + {\left(\frac{3}{2}\right)}^{2} = 1 + \frac{9}{4} = \frac{13}{4.}$
$\sec \theta = \pm \frac{\sqrt{13}}{2.}$
Since, $\theta$ lies in ${Q}_{I I I} , \sec \theta = - \frac{\sqrt{13}}{2.}$