Prove that: $$\cot 20^\circ - \cot 40^\circ + \cot 80^\circ = \sqrt{3}$$
What I have learnt in trigonometry so far: Trigonometric ratios and their graphs, formulas of ratios for compound angles, sum and product formulas of ratios (eg. $2 \sin A \cos B = \sin (A+B) + \sin (A-B)$ ).
An answer using the things I have learnt would be appreciated. I tried this multiple times by trying to simplify to tan and then using the expression for $\tan (A+B)$, but it didn't work out.
EDIT: My attempt in detail:
$$\cot 20^\circ - \cot 40^\circ + \cot 80^\circ = [(\tan 20^\circ + \tan 80^\circ) / \tan 20^\circ \tan 80^\circ ] - 1 / \tan 40^\circ$$
After this I further tried to get it in some form of ratio with which I could get angles like $ 30^\circ$ or $ 60^\circ $, but I was unable to do it. There were two other proofs prior to this one, which seemed to use the same strategy.
Source: Challenge and Thrill of Pre-College Mathematics
Here is a solution using the formula for the sum of cosines of compound angles: $$\cos{(A+B)} \; + \; \cos{(A-B)}=2\cos{A}\cos{B}$$
An intuitive point here is that we see double angles (20, 40 ,80) so we can use formulae for $\sin{2A}$ and $\cos{2A}$. The trick is to work with two terms at a time, as marked with colors below. We also note that $\sin{(90-A)}=\cos{A}$ .
Alright, let's start with the two leftmost terms: $$ \begin{align} &\quad \; \color{blue}{\cot20 \; - \; \cot40} \; + \cot80 \\[2ex] &= \color{blue}{\frac{\cos20}{\sin20} \; - \; \frac{\cos40}{\sin40}} \; + \; \frac{\cos80}{\sin80} \tag1 \\[2ex] &= \color{blue}{\frac{\cos20}{\sin20} \; - \; \frac{2\cos^{2}20-1}{2\sin20\cos20}} \; + \; \frac{\cos80}{\sin80} \tag2 \\[2ex] &= \color{blue}{\frac{1}{\sin40}} \; + \; \frac{\cos80}{2\sin40\cos40} \tag3 \\[2ex] &= \frac{2\cos40 \; + \; \cos80}{2\sin40\cos40} \tag4 \\[2ex] &= \frac{\cos40 \; + \; \color{green}{\cos40 \; + \; \cos80}}{\sin80} \tag5 \\[2ex] &= \frac{\cos40 \; + \; \color{green}{2\cos60\cos20}}{\sin80} \tag6 \\[2ex] &= \frac{\cos40 \; + \; \color{green}{\cos20}}{\cos10} \tag7 \\[2ex] &= \frac{2\cos30\cos10}{\cos10} \tag8 \\[2ex] &= 2\cos30 = \sqrt3 \tag9 \\[2ex] \end{align} $$