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, 
$$\sin7.5 = \sqrt{\dfrac{1-\cos 15} {2}}$$
$$\cos7.5 = \sqrt{\dfrac{1+\cos 15} {2}} $$
$$\implies \cot 7.5 = \sqrt{\dfrac{2\sqrt2 + \sqrt3 + 1} {2\sqrt2 - \sqrt3 + 1}} $$

using half angle identities,
$$\cot(x)=\frac{1}{\tan(x)}=\frac{\sin(2x)}{1-\cos(2x)}=\frac{\left(\frac{1-\cos(4x)}{2}\right)^{\frac{1}{2}}}{1-\left(\frac{1+\cos(4x)}{2}\right)^\frac{1}{2}}$$
with $x=7.5^o=\frac{\pi}{24}$ and $\cos(\frac{4\pi}{24})=\frac{\sqrt{3}}{2}$ we can substitute and simplify by multiplying through by $\sqrt{2}$ twice,
$$
\cot(\frac{\pi}{24})
=
\frac
{
  \frac
    { \left( 1-\frac{\sqrt{3}}{2} \right) ^{\frac{1}{2}} }
    { ( {2} )^{\frac{1}{2}} }
  }
  {
    1- \frac
        { \left( 1+\frac{\sqrt{3}}{2} \right) ^{\frac{1}{2}} }
        { ( {2} )^{\frac{1}{2}}
  }
}
=
\frac
{
  \frac
    { ( 2-\sqrt{3} ) ^{\frac{1}{2}} }
    { ( {2} ) ^{\frac{1}{2}} }
  }
{
  \sqrt{2} - \frac
               {(2+\sqrt{3}) ^{\frac{1}{2}}}
               {{(2)}^{\frac{1}{2}}}
}
=
\frac
{
  ( 2-\sqrt{3} ) ^{\frac{1}{2}}
}
{
  2 - (2+\sqrt{3}) ^{\frac{1}{2}}
}
$$
to simplify from here, we could look to 'complete the square' under the radicals (leaving NO remainder) in order for the the powers $2$ and $1/2$ to multiply and cancel
look for $a, b$ such that $(a + b)^2 = 2+\sqrt{3}$ ideally (and similarly for the $2-\sqrt{3}$ radical), but note that any scaler multiple of this RHS would do as this multiple can be either factored out, i.e. completing for $k(2-\sqrt{3}) = (a-b)^2$ works if we replace $(2-\sqrt{3})$ with $\left(\frac{1}{\sqrt{k}}\right)^2(a-b)^2$ instead, or alternatively pulled in from outside of the radical, i.e. by multiplying the main equation through by the factor $\sqrt{k}$
this is the weakest part of the solution, as only trial and error led me to $k=2, a=1, b=\pm\sqrt{3}$
$$
\cot(\frac{\pi}{24})
=
\frac
  {\left(\left(\frac{1}{\sqrt{2}}\right)^2(1-\sqrt{3})^2\right)^\frac{1}{2}}
  {2-\left(\left(\frac{1}{\sqrt{2}}\right)^2(1+\sqrt{3})^2\right)^\frac{1}{2}}
=
\frac
  {\frac{1}{\sqrt{2}}|1-\sqrt{3}|}
  {2-\frac{1}{\sqrt{2}}(1+\sqrt{3})}
=
\frac
  {|1-\sqrt{3}|}
  {2\sqrt{2}-(1+\sqrt{3})}
$$
to simplify the (disgusting) modulus (which exists to maintain the exact radical solution), multiply through by its conjugate. being very careful with signs (positive $|1-\sqrt{3}|$ scaled by positive $(1+\sqrt{3})$ is positive) we have,
$$
\cot(\frac{\pi}{24})
=
\frac
{
  |1-\sqrt{3}|(1+\sqrt{3})
}
{
  2\sqrt{2}(1+\sqrt{3})-(1+\sqrt{3})^2
}
=\frac{2}{2\sqrt{2}+2\sqrt{6}-(4+2\sqrt{3})}
$$
finally, multiplying through by $\frac{1}{2}(\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6})$, the denominator simplifies to $1$, and we're left with your required solution
