# Proving $\cot 20^\circ - \cot 40^\circ + \cot 80^\circ = \sqrt{3}$

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

• – lhf
Jun 17, 2021 at 13:00
• math.stackexchange.com/questions/455070/… Jun 17, 2021 at 13:02
• @MathLover I have edited my post to include the basics of what I tried, all further attempts stemmed from it. Jun 17, 2021 at 13:06
• labbhattacharjee's linked answer has the correct strategy. Note that $\cot 40^\circ = -\cot 140^\circ$ Jun 17, 2021 at 13:09
• Small correction: in your example formula for compound angles, please change the left hand side to $2 \sin A \cos B$ . Jun 18, 2021 at 0:48

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}