The problem asks us to find the value of $\tan(\frac{2\pi}{3})$. And the second problem asks to simplify $\frac{2\tan(\frac{\pi}{12})}{1-\tan^2(\frac{\pi}{12})}$.

1. Problem Description

The problem asks us to find the value of $\tan(\frac{2\pi}{3})$.

And the second problem asks to simplify $\frac{2\tan(\frac{\pi}{12})}{1-\tan^2(\frac{\pi}{12})}$.

2. Solution Steps

First, let's evaluate $\tan(\frac{2\pi}{3})$. We know that $\frac{2\pi}{3}$ is in the second quadrant.

We can write $\frac{2\pi}{3}$ as $\pi - \frac{\pi}{3}$.

The tangent function is negative in the second quadrant.

$\tan(\frac{2\pi}{3}) = \tan(\pi - \frac{\pi}{3}) = -\tan(\frac{\pi}{3}) = -\sqrt{3}$.

Next, let's simplify $\frac{2\tan(\frac{\pi}{12})}{1-\tan^2(\frac{\pi}{12})}$.

Recall the double angle formula for tangent:

$\tan(2x) = \frac{2\tan(x)}{1 - \tan^2(x)}$.

Using this formula, we can simplify the given expression:

$\frac{2\tan(\frac{\pi}{12})}{1-\tan^2(\frac{\pi}{12})} = \tan(2 \cdot \frac{\pi}{12}) = \tan(\frac{\pi}{6})$.

We know that $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$.

3. Final Answer

The value of $\tan(\frac{2\pi}{3})$ is $-\sqrt{3}$.

The simplified form of $\frac{2\tan(\frac{\pi}{12})}{1-\tan^2(\frac{\pi}{12})}$ is $\frac{1}{\sqrt{3}}$.