Step 1: Use identity: \[ \tan(2A) = \frac{2 \tan A}{1 - \tan^2 A} \quad \text{But this is different, let's evaluate directly.} \]
Step 2: Substitute values. \[ \tan 30^\circ = \frac{1}{\sqrt{3}} \Rightarrow \tan^2 30^\circ = \frac{1}{3} \] So, \[ \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}} = \frac{2}{\sqrt{3}} \cdot \frac{3}{4} = \frac{6}{4\sqrt{3}} = \frac{3}{2\sqrt{3}} \] Now, \[ \sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \Rightarrow \frac{3}{2\sqrt{3}} = \sin 60^\circ \]