<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 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:
$$7^{\circ}30' = 15^{\circ}\cdot \frac{1}{2} =30^{\circ}\cdot \frac{1}{2}\cdot \frac{1}{2} $$
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