The problem asks us to evaluate $\tan(\frac{7\pi}{3})$.

1. Problem Description

The problem asks us to evaluate $\tan(\frac{7\pi}{3})$.

2. Solution Steps

First, we can simplify the argument of the tangent function. We want to find an angle coterminal with $\frac{7\pi}{3}$ that lies between $0$ and $2\pi$. We can do this by subtracting multiples of $2\pi$ from $\frac{7\pi}{3}$:

$\frac{7\pi}{3} - 2\pi = \frac{7\pi}{3} - \frac{6\pi}{3} = \frac{\pi}{3}$.

Therefore, $\tan(\frac{7\pi}{3}) = \tan(\frac{\pi}{3})$.

The angle $\frac{\pi}{3}$ radians is $60^\circ$.

$\tan(\frac{\pi}{3}) = \frac{\sin(\frac{\pi}{3})}{\cos(\frac{\pi}{3})}$.

We know that $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ and $\cos(\frac{\pi}{3}) = \frac{1}{2}$.

Therefore,

$\tan(\frac{\pi}{3}) = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{\sqrt{3}}{2} \cdot \frac{2}{1} = \sqrt{3}$.

3. Final Answer

$\sqrt{3}$