<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] 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. 
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