Prove that: $ \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$

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 $$\cot(7.5^\circ) = \sqrt{8+4\sqrt{3}} +2+\sqrt{3} = \sqrt{2}+\sqrt{6} + \sqrt{4}+\sqrt{3}$$

$$\text{As } \cot x =\frac{\cos x}{\sin x}$$ $$ =\frac{2\cos^2x}{2\sin x\cos x}(\text{ multiplying the numerator & the denominator by }2\cos7\frac12 ^\circ)$$

$$=\frac{1+\cos2x}{\sin2x}(\text{using }\sin2A=2\sin A\cos A,\cos2A=2\cos^2A-1$$

$$ \cot7\frac12 ^\circ =\frac{1+\cos15^\circ}{\sin15^\circ}$$

$\cos15^\circ=\cos(45-30)^\circ=\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ=\frac{\sqrt3+1}{2\sqrt2}$

$\sin15^\circ=\sin(45-30)^\circ=\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ=\frac{\sqrt3-1}{2\sqrt2}$

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 $2:$

$$\implies \frac{1+\cos15^\circ}{\sin15^\circ}=\frac{1+\frac{\sqrt3+1}{2\sqrt2}}{\frac{\sqrt3-1}{2\sqrt2}}=\frac{2\sqrt2+\sqrt3+1}{\sqrt3-1}=\frac{(2\sqrt2+\sqrt3+1)(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}(\text{ rationalizing the denominator })$$

$$=\frac{2\sqrt6+4+2\sqrt3+2\sqrt2}2$$

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 $\sqrt{2\sqrt{2}−\sqrt{3}+1}$.