Answer
$\cos\theta$ = $\frac{1}{3}$
$\sin\theta$ = - $\frac{2\sqrt 2}{3}$
$\tan\theta$ = $-2\sqrt 2$
Work Step by Step
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. $\tan\theta$ = $-2\sqrt 2$
