## Trigonometry 7th Edition

$\sec\theta$ = - $\frac{25}{24}$
We know from third Pythagorean identity that- $\sec\theta$ = ± $\sqrt (1+\tan^{2}\theta)$ As $\cos\theta\lt0$, i.e. $\cos\theta$ is negative and $\tan\theta$ is positive, hence $\theta$ terminates in Q III, Therefore $\sec\theta$ will be negative- $\sec\theta$ = - $\sqrt (1+\tan^{2}\theta)$ substitute the given value of $\tan\theta$- $\sec\theta$ = - $\sqrt (1+(\frac{7}{24})^{2})$ $\sec\theta$= - $\sqrt (1+\frac{49}{576})$ $\sec\theta$ = - $\sqrt (\frac{576+49}{576})$ = $\sqrt (\frac{625}{576})$ $\sec\theta$ = - $\frac{25}{24}$