<s>[INST] 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.
 [/INST] 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.
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