TITLE: Prove that: $ \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$
QUESTION [23 upvotes]: How to prove the following trignometric identity?
$$ \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$$
Using half angle formulas, I am getting a number for $\cot7\frac12 ^\circ $, but I don't know how to show it to equal the number $\sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$.
I would however like to learn the technique of dealing with surds such as these, especially in trignometric problems as I have a lot of similar problems and I don't have a clue as to how to deal with those.
Hints please!
EDIT:
What I have done using half angles is this: (and please note, for convenience, I am dropping the degree symbols. The angles here are in degrees however).
I know that
$$ \cos 15 = \dfrac{\sqrt3+1}{2\sqrt2}$$
So, 
$$\sin7.5 = \sqrt{\dfrac{1-\cos 15} {2}}$$
$$\cos7.5 = \sqrt{\dfrac{1+\cos 15} {2}} $$
$$\implies \cot 7.5 = \sqrt{\dfrac{2\sqrt2 + \sqrt3 + 1} {2\sqrt2 - \sqrt3 + 1}} $$

REPLY [2 votes]: Alternatively:
$$\cot{(7.5)}=\frac{\cos{(7.5)}}{\sin{(7.5)}}=\frac{2\cos^2{(7.5)}}{\sin{(15)}}=\frac{1+\cos{(15)}}{\sin{(15)}}=$$
$$\frac{2(\cos{(15)}+\cos^2{(15)})}{\sin{(30)}}=4\left(\cos{(45-30)}+\frac{1+\cos{(30)}}{2}\right)=4\left(\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot \frac{1}{2}+\frac12+\frac{\sqrt{3}}{4}\right)=$$
$$\sqrt{6}+\sqrt{2}+\sqrt{4}+\sqrt{3}.$$