Prove: $\tan\frac{\pi}{24}=2\sqrt{2+\sqrt{3}}-\sqrt{3}-2$ 
How to prove that
  $$\tan\frac{\pi}{24}=2\sqrt{2+\sqrt{3}}-\sqrt{3}-2$$

I get $$\tan\frac{\pi}{24}=\sqrt\frac{2\sqrt{2}-\sqrt{3}-1}{ 2\sqrt{2}+\sqrt{3}+1}$$ but i can't transform it.

$tan2\theta = 2tan\theta/(1-tan\theta^2)$
Let $tan\pi/12 = a, tan\pi/24 = x$, use formula, we get 
$(x = -1 - \sqrt{1+a^2})/a = (-1+ \sqrt{2-\sqrt{3}})/(2-\sqrt{3}) = -1 / (2-\sqrt{3}) + 2 / \sqrt{(2-\sqrt{3})} 
= -2 - \sqrt{3} + 2\sqrt{(2+\sqrt{3})}$
