# Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 480: 25

$\dfrac{\sqrt 2}{4}$

#### 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, $\tan (\sin^{-1} \dfrac{1}{3})=\tan \theta =\dfrac{1}{2\sqrt 2}=\dfrac{\sqrt 2}{4}$

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