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,

$$\sin7.5 = \sqrt{\dfrac{1-\cos 15} {2}}$$ $$\cos7.5 = \sqrt{\dfrac{1+\cos 15} {2}} $$

$$\implies \cot 7.5 = \sqrt{\dfrac{2\sqrt2 + \sqrt3 + 1} {2\sqrt2 - \sqrt3 + 1}} $$

Solution 1:

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

Solution 2:

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

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

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

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

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

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 $2:$

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

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