# If the tan(x)=3/2 in quadrant III, what is the trigonometric value of the sec(x)?

Apr 6, 2016

$- \frac{\sqrt{13}}{2}$
$1 + {\tan}^{2} x = {\sec}^{2} x$
${\sec}^{2} x = 1 + \frac{9}{4} = \frac{13}{4}$
$\sec x = \pm \frac{\sqrt{13}}{2}$
$\sec x = - \frac{\sqrt{13}}{2}$