1. The problem is to evaluate $\tan 240^\circ$. 2. Recall that the tangent function is periodic with period $180^\circ$, and $\tan(\theta) = \frac{\sin \theta}{\cos \theta}$. 3. The angle $240^\circ$ lies in the third quadrant where both sine and cosine are negative, so tangent is positive. 4. We can write $240^\circ = 180^\circ + 60^\circ$. 5. Using the tangent addition formula for angles in the third quadrant: $\tan(180^\circ + \alpha) = \tan \alpha$. 6. Therefore, $\tan 240^\circ = \tan 60^\circ$. 7. We know $\tan 60^\circ = \sqrt{3}$. 8. Hence, $\tan 240^\circ = \sqrt{3}$. Final answer: $\sqrt{3}$