#### Answer

$$\frac{2\tan15^\circ}{1-\tan^215^\circ}=\frac{\sqrt3}{3}$$
2 is matched with E.

#### Work Step by Step

$$X=\frac{2\tan15^\circ}{1-\tan^215^\circ}$$
Recall the double-angle identities:
$$\frac{2\tan A}{1-\tan^2A}=\tan2A$$
Here $X$ is like the form of $\frac{2\tan A}{1-\tan^2A}$ with $A=15^\circ$. Therefore,
$$X==\frac{2\tan15^\circ}{1-\tan^215^\circ}=\tan(2\times15^\circ)$$
$$X=\tan30^\circ$$
$$X=\frac{1}{\sqrt3}=\frac{\sqrt3}{3}$$
So, $$\frac{2\tan15^\circ}{1-\tan^215^\circ}=\frac{\sqrt3}{3}$$
We pick E here.