Answer
$$\frac{2\tan15^\circ}{1-\tan^215^\circ}=\frac{\sqrt3}{3}$$
Work Step by Step
$$X=\frac{2\tan15^\circ}{1-\tan^215^\circ}$$
- From Double-Angle Identity for tangent: $$\tan2A=\frac{2\tan A}{1-\tan^2A}$$
So if we replace the above identity with $A=15^\circ$ as in $X$, we get
$$X=\tan(2\times15^\circ)$$
$$X=\tan30^\circ$$
$$X=\frac{1}{\sqrt3}=\frac{\sqrt3}{3}$$
Therefore, $$\frac{2\tan15^\circ}{1-\tan^215^\circ}=\frac{\sqrt3}{3}$$
