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, 
$$\sin7.5 = \sqrt{\dfrac{1-\cos 15} {2}}$$
$$\cos7.5 = \sqrt{\dfrac{1+\cos 15} {2}} $$
$$\implies \cot 7.5 = \sqrt{\dfrac{2\sqrt2 + \sqrt3 + 1} {2\sqrt2 - \sqrt3 + 1}} $$
 A: Start from $\displaystyle\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{2\tan{15^\circ}}{1-\tan^2{15^\circ}}$.
If $x=\tan{15^\circ}$, then $\displaystyle\tan 15^\circ = x = \frac{2\tan{(\frac{15}{2})^\circ}}{1-\tan^2{(\frac{15}{2})^\circ}}$.
If $y=\tan{(\frac{15}{2})^\circ}$, then $x=\frac{2y}{1-y^2}$. Hence
$\displaystyle\frac{1}{\sqrt{3}} = \frac{2(\frac{2y}{1-y^2})}{1-(\frac{2y}{1-y^2})^2}$.
Simplify the above equation and solve for $y$, then find the reciprocal to find $\cot{(\frac{15}{2})^\circ}$.
EDIT: To simplify your surd, try to multiply both the numerator and denominator by $\sqrt{2\sqrt{2}−\sqrt{3}+1}$.
A: Note that if $\cot(x) = c$, $$\cot(4x) = \frac{1-6c^2+c^4}{4 c^3 - 4 c}$$
In this case $\cot(4x) = \sqrt{3}$.  thus you want
$$ c^4 - 6 c^2 + 1 - (4 c^3 - 4 c) \sqrt{3} = 0 $$
The quartic happens to factor as
$$ (c^2 + (4-2 \sqrt{3}) c - 1)(c^2 + (-4 - 2 \sqrt{3}) c - 1)$$
Use numerical approximation to see which quadratic factor you want to be $0$, and
solve.  You may also find it useful to note that $4 + 2 \sqrt{3} = (1+\sqrt{3})^2 $.
A: $$\cot7.5^{\circ}=\sqrt{\frac{1+\cos15^{\circ}}{1-\cos15^{\circ}}}=\sqrt{\frac{1+\sqrt{\frac{1+\frac{\sqrt3}{2}}{2}}}{1-\sqrt{\frac{1+\frac{\sqrt3}{2}}{2}}}}=\sqrt{\frac{\sqrt8+\sqrt3+1}{\sqrt8-\sqrt3-1}}=$$
$$=\frac{\sqrt{8-4-2\sqrt3}}{\sqrt8-\sqrt3-1}=\frac{\sqrt3-1}{\sqrt8-\sqrt3-1}=\frac{(\sqrt3-1)(2\sqrt2+\sqrt3+1))}{8-4-2\sqrt3}=$$
$$=\frac{2\sqrt2+\sqrt3+1}{\sqrt3-1}=\frac{1}{2}(\sqrt3+1)(2\sqrt2+\sqrt3+1)=\sqrt2+\sqrt3+\sqrt4+\sqrt6.$$
Done!
A: Alternatively:
$$\cot{(7.5)}=\frac{\cos{(7.5)}}{\sin{(7.5)}}=\frac{2\cos^2{(7.5)}}{\sin{(15)}}=\frac{1+\cos{(15)}}{\sin{(15)}}=$$
$$\frac{2(\cos{(15)}+\cos^2{(15)})}{\sin{(30)}}=4\left(\cos{(45-30)}+\frac{1+\cos{(30)}}{2}\right)=4\left(\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot \frac{1}{2}+\frac12+\frac{\sqrt{3}}{4}\right)=$$
$$\sqrt{6}+\sqrt{2}+\sqrt{4}+\sqrt{3}.$$
A: Here is an elementary (almost without words) proof that does not rely explicitly on half-angle or double-angle formulas. What is used is the fact that 
$\cot(30^\circ) = \sqrt{3}$, the exterior angle theorem, isosceles triangle theorems and Pythagorean theorem of Euclidean geometry, and the fact that 
$8 + 4\sqrt{3} = \left(\sqrt{2}+\sqrt{6}\right)^2$ (cf. Robert Israel's answer).
The crude, not-to-scale, diagram below is, I hope, self-explanatory especially if
you start from the right side and work your way to the left.
The length of the base of the triangle is
$$\cot(7.5^\circ) = \sqrt{8+4\sqrt{3}} +2+\sqrt{3} = \sqrt{2}+\sqrt{6} + \sqrt{4}+\sqrt{3}$$

A: $$\text{As } \cot x =\frac{\cos x}{\sin x}$$
$$ =\frac{2\cos^2x}{2\sin x\cos x}(\text{ multiplying the numerator & the denominator by }2\cos7\frac12 ^\circ)$$
$$=\frac{1+\cos2x}{\sin2x}(\text{using }\sin2A=2\sin A\cos A,\cos2A=2\cos^2A-1$$
$$ \cot7\frac12 ^\circ =\frac{1+\cos15^\circ}{\sin15^\circ}$$
$\cos15^\circ=\cos(45-30)^\circ=\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ=\frac{\sqrt3+1}{2\sqrt2}$
$\sin15^\circ=\sin(45-30)^\circ=\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ=\frac{\sqrt3-1}{2\sqrt2}$
Method $1:$
$$\frac{1+\cos15^\circ}{\sin15^\circ}=\csc15^\circ+\cot15^\circ$$
$$\cot15^\circ=\frac{\cos15^\circ}{\sin15^\circ}=\frac{\sqrt3+1}{\sqrt3-1}=\frac{(\sqrt3+1)^2}{(\sqrt3-1)(\sqrt3+1)}=2+\sqrt3$$
$$\csc15^\circ=\frac{2\sqrt2}{\sqrt3-1}=\frac{2\sqrt2(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}=\sqrt2(\sqrt3+1)=\sqrt6+\sqrt2$$
Method $2:$
$$\implies \frac{1+\cos15^\circ}{\sin15^\circ}=\frac{1+\frac{\sqrt3+1}{2\sqrt2}}{\frac{\sqrt3-1}{2\sqrt2}}=\frac{2\sqrt2+\sqrt3+1}{\sqrt3-1}=\frac{(2\sqrt2+\sqrt3+1)(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}(\text{ rationalizing the denominator  })$$
$$=\frac{2\sqrt6+4+2\sqrt3+2\sqrt2}2$$
A: \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}
A: 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
