Simplifying trig expression I have 
$$\frac{\tan{15^\circ}}{1-\tan{15^\circ}^2}$$
and need to simplify it. The only equation I have that is even close to a match for it is $2\frac{\tan{15^\circ}}{1-\tan{15^\circ}^2}$. But the numerator isn't right with the $\times 2$. The answer should be $\frac{\sqrt{3}}{6}$. What do I do with this?

$$
2 \frac{\tan(15^{\circ})}{1-\tan(15 ^{\circ})^2} = \tan(30) = \frac{1}{\sqrt{3}} \tag{1}
$$ 
because:
$$
 \frac{\tan(\alpha) + \tan(\beta)}{1-\tan(\alpha)\tan(\beta)} = \tan(\alpha+\beta) \tag{2}
$$ 
Following equation (1):
$$
\frac{\tan(15^{\circ})}{1-\tan(15 ^{\circ})^2} = \frac{1}{2}\tan(30) = \frac{1}{2\sqrt{3}}
$$
And finally we employ a very useful method in dealing with roots, multiplying by $1$:
$$
\frac{1}{2\sqrt{3}}=\frac{1}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{6}$$  
