Prove that $\tan 7°30' = \sqrt {6} - \sqrt {3} + \sqrt {2} - 2$ Prove that:
$$\tan 7°30' = \sqrt {6} - \sqrt {3} + \sqrt {2} - 2$$
My Work:
I guess that I have to use the formula :
$$\tan A = \frac {2 \tan(\frac {A}{2})}{1-\tan^2 (\frac {A}{2})}$$
But, I am not being able to use it. Please help me.

Hint
$$\tan{(15^{\circ})}=2-\sqrt{3}$$
and let$x=\tan{7.5^{\circ}}$.we have
$$\dfrac{2x}{1-x^2}=2-\sqrt{3}\Longrightarrow (2-\sqrt{3})x^2+2x-(2-\sqrt{3})=0$$
so we have
$$x^2+2(2+\sqrt{3})x-1=0$$
then we have
$$(x+2+\sqrt{3})^2=1+(2+\sqrt{3})^2=8+4\sqrt{3}=2(\sqrt{3}+1)^2$$
then we have
$$x=\sqrt{2}(\sqrt{3}+1)-2-\sqrt{3}=\sqrt{6}+\sqrt{2}-2-\sqrt{3}$$
