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.
 A: Hint: We have $$\tan 7.5^\circ =\frac {\sin 7.5^\circ}{\cos 7.5^\circ} =\frac {2\sin^2 7.5^\circ}{2\cos 7.5^\circ \sin 7.5^\circ} =\frac {1-\cos 15^\circ}{\sin 15^\circ} =\frac {1- \cos (45^\circ -30^\circ)}{\sin (45^\circ -30^\circ)} $$ 
Can you take it from here? Hope it helps. 
A: $\cos 15^{\circ}=\cos(45^{\circ}-30^{\circ})=\cos45^{\circ}\cos30^{\circ}+\sin30^{\circ}\sin45^{\circ}=\frac{\sqrt6+\sqrt2}{4}$
$$\tan \frac{15^{\circ}}2=\sqrt{\frac{1-\cos15^{\circ}}{1+\cos15^{\circ}}}$$
A: You may prove that identity in a very geometric flavour, i.e. by bisecting twice a $30^\circ$ angle, through the Pythagorean theorem and the bisector theorem. It is best to keep the expressions of the involved lengths as simple as possible during the process. So, let we consider a triangle $BAE$ with $BA=\sqrt{3}, AE=2, EB=1$. Let $AD$ be the bisector of $\widehat{BAE}$ and $AC$ be the bisector of $\widehat{BAD}$.

By the bisector theorem, we have:
$$ BD = \frac{\sqrt{3}}{2+\sqrt{3}} = 2\sqrt{3}-3. \tag{1}$$
By the Pythagorean theorem, it follows that:
$$ AD^2 = 3+(2\sqrt{3}-3)^2 = 24-12\sqrt{3} = 6(\sqrt{3}-1)^2 \tag{2} $$
hence $AD=3\sqrt{2}-\sqrt{6}$. By the bisector theorem again,
$$ BC = BD\cdot\frac{AB}{AB+AD},\quad \frac{BC}{BA} = \frac{BD}{BA+AD} = \frac{2\sqrt{3}-3}{\sqrt{3}+3\sqrt{2}-\sqrt{6}}\tag{3} $$
so:
$$ \tan(7^\circ 30') = \frac{2-\sqrt{3}}{1+\sqrt{6}-\sqrt{2}}\tag{4}$$
and the claim turns out to be equivalent to:
$$ (1+\sqrt{6}-\sqrt{2})(\sqrt{6}-\sqrt{3}+\sqrt{2}-2)=(2-\sqrt{3})\tag{5} $$
that is tedious but straightforward to check.
A: 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
$$x=\sqrt{2}(\sqrt{3}+1)-2-\sqrt{3}=\sqrt{6}+\sqrt{2}-2-\sqrt{3}$$
A: Hint 1:
For $x\in [0,\frac{\pi}{2}]$:
$$\tan \frac{x}{2} = \frac{\sin x}{1+\cos x}$$
$$\cos \frac{x}{2} = \sqrt{\frac{1+\cos x}{2}}$$
$$\sin \frac{x}{2} = \sqrt{\frac{1-\cos x}{2}}$$
Hint 2:
$$7^{\circ}30' = 15^{\circ}\cdot \frac{1}{2} =30^{\circ}\cdot \frac{1}{2}\cdot \frac{1}{2} $$
