Prove that: $ \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$ 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, 
$$\sin7.5 = \sqrt{\dfrac{1-\cos 15} {2}}$$
$$\cos7.5 = \sqrt{\dfrac{1+\cos 15} {2}} $$
$$\implies \cot 7.5 = \sqrt{\dfrac{2\sqrt2 + \sqrt3 + 1} {2\sqrt2 - \sqrt3 + 1}} $$

\begin{align}
& \cot({7.5^\circ}) = \cot\left(\frac{45^\circ-30^\circ}{2}\right)
= \frac{\cos(45^\circ) + \cos(30^\circ)}{\sin(45^\circ)-\sin(30^\circ)} \\[10pt]
= {} & \frac{\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2} - \frac{1}{2}} =\frac{ 2 + \sqrt{6} }{ 2 - \sqrt{2}} 
=\sqrt{6} + \sqrt{3} + \sqrt{2} + 2
\end{align}
