Answer
The exact value of $\frac{2\tan \frac{\pi }{8}}{1-{{\tan }^{2}}\frac{\pi }{8}}$ is $1$.
Work Step by Step
Recall the given expression.
$\tan 2\theta =\frac{2\tan \theta }{1-{{\tan }^{2}}\theta }$
Apply the given expression.
$\begin{align}
& \frac{2\tan \frac{\pi }{8}}{1-{{\tan }^{2}}\frac{\pi }{8}}=\tan 2\left( \frac{\pi }{8} \right) \\
& =\tan \left( \frac{\pi }{4} \right) \\
& =1
\end{align}$
Therefore, the exact value of $\frac{2\tan \frac{\pi }{8}}{1-{{\tan }^{2}}\frac{\pi }{8}}$ is $1$.
