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