Answer
$\frac{1}{\sqrt {3}}$
Work Step by Step
Let $\sin^{-1}(\frac{1}{2})=\theta$
Then, $\sin \theta=\frac{1}{2}=\sin \frac{\pi}{6}$
$⇒\theta=\frac{\pi}{6}$.
Therefore, $\tan(\sin^{-1}\frac{1}{2})= \tan(\frac{\pi}{6})= \frac{1}{\sqrt 3}$
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