Answer
$\sec\theta$ = - $\frac{17}{15}$
Work Step by Step
We know from second Pythagorean identity that-
$\sec\theta$ = ± $\sqrt (1+\tan^{2}\theta)$
As $\theta$ terminates in Q III, Therefore $\sec\theta$ will be negative-
$\sec\theta$ = - $\sqrt (1+\tan^{2}\theta)$
substitute the given value of $\tan\theta$-
$\sec\theta$ = - $\sqrt (1+(\frac{8}{15})^{2})$
$\sec\theta$= - $\sqrt (1+\frac{64}{225})$
$\sec\theta$ = - $\sqrt (\frac{225+64}{225})$ = $\sqrt (\frac{289}{225})$
$\sec\theta$ = - $\frac{17}{15}$
