## Trigonometry 7th Edition

$\cos\theta$ = $\frac{1}{3}$ $\sin\theta$ = - $\frac{2\sqrt 2}{3}$ $\tan\theta$ = $-2\sqrt 2$
Given $\sec\theta$ = 3 Using reciprocal identity- $\cos\theta$ = $\frac{1}{\sec\theta}$ $\cos\theta$ = $\frac{1}{3}$ From first Pythagorean identity- $\sin\theta$ = ± $\sqrt {1-\cos^{2}\theta}$ As $\theta$ terminates in Q IV, therefore $\sin\theta$ will be negative, hence- $\sin\theta$ = - $\sqrt {1-\cos^{2}\theta}$ = - $\sqrt {1-(\frac{1}{3})^{2}}$ = -$\sqrt {1 - \frac{1}{9}}$ = -$\sqrt {\frac{9-1}{9}}$ = -$\sqrt {\frac{8}{9}}$ i.e. $\sin\theta$ = - $\frac{2\sqrt 2}{3}$ From second Pythagorean identity, $\tan\theta$ = ± $\sqrt {\sec^{2}\theta - 1}$ As $\theta$ terminates in Q IV, therefore $\tan\theta$ will be negative, hence- $\tan\theta$ = - $\sqrt {\sec^{2}\theta - 1}$ = - $\sqrt {3^{2} - 1}$ = - $\sqrt {9 - 1}$ = - $\sqrt {8}$ i.e. $\tan\theta$ = $-2\sqrt 2$