Answer
$\dfrac{\sqrt 2}{4}$
Work Step by Step
Let us consider: $\theta =\sin^{-1} \dfrac{1}{3}$
This can be written as: $\sin \theta =\dfrac{1}{3}$
The trigonometric functions can be written as below:
$\sin \theta=\dfrac{Opposite}{Hypotenuse}$
$\cos \theta=\dfrac{Adjacent}{Hypotenuse}$
$\tan \theta=\dfrac{Opposite}{Adjacent}$
So, we have: the opposite side is $1$ and the hypotenuse is $3$.
Thus, by the Pythagorean Theorem, the opposite side is $=\sqrt {3^2-1^2}=\sqrt 8=2\sqrt 2$
Since, $\tan \theta=\dfrac{Opposite}{Adjacent}$
Therefore, $\tan (\sin^{-1} \dfrac{1}{3})=\tan \theta =\dfrac{1}{2\sqrt 2}=\dfrac{\sqrt 2}{4}$
