## Trigonometry 7th Edition

$\tan\theta$ = $\frac{1}{2\sqrt 2}$
To find $\tan\theta$, we will calculate $\cos\theta$ first- We know from first Pythagorean identity that- $\cos\theta$ = ± $\sqrt (1-\sin^{2}\theta)$ As $\theta$ terminates in Q I, Therefore $\cos\theta$ will be positive- $\cos\theta$ = $\sqrt (1-\sin^{2}\theta)$ substitute the given value of $\sin\theta$- $\cos\theta$ = $\sqrt (1-(\frac{1}{3})^{2})$ $\cos\theta$ = $\sqrt (1-\frac{1}{9})$ $\cos\theta$ = $\sqrt (\frac{9 - 1}{9})$ = $\sqrt (\frac{8}{9})$ $\cos\theta$ = $\frac{2\sqrt 2}{3}$ As $\theta$ terminates in Q I, Therefore $\tan\theta$ will also be positive- By ratio identity- $\tan\theta$ = $\frac{\sin\theta}{\cos\theta}$ Substituting the values of $\sin\theta$ and $\cos\theta$- $\tan\theta$ = $\frac{1/3}{2\sqrt 2/3}$ = $\frac{1}{2\sqrt 2}$