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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

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    $\begingroup$ See math.stackexchange.com/questions/770391/… $\endgroup$
    – lhf
    Jun 17, 2021 at 13:00
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    $\begingroup$ math.stackexchange.com/questions/455070/… $\endgroup$ Jun 17, 2021 at 13:02
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    $\begingroup$ @MathLover I have edited my post to include the basics of what I tried, all further attempts stemmed from it. $\endgroup$
    – Aether162
    Jun 17, 2021 at 13:06
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    $\begingroup$ labbhattacharjee's linked answer has the correct strategy. Note that $\cot 40^\circ = -\cot 140^\circ$ $\endgroup$ Jun 17, 2021 at 13:09
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    $\begingroup$ Small correction: in your example formula for compound angles, please change the left hand side to $2 \sin A \cos B$ . $\endgroup$
    – Saeed
    Jun 18, 2021 at 0:48

1 Answer 1

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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: $$ \begin{align} &\quad \; \color{blue}{\cot20 \; - \; \cot40} \; + \cot80 \\[2ex] &= \color{blue}{\frac{\cos20}{\sin20} \; - \; \frac{\cos40}{\sin40}} \; + \; \frac{\cos80}{\sin80} \tag1 \\[2ex] &= \color{blue}{\frac{\cos20}{\sin20} \; - \; \frac{2\cos^{2}20-1}{2\sin20\cos20}} \; + \; \frac{\cos80}{\sin80} \tag2 \\[2ex] &= \color{blue}{\frac{1}{\sin40}} \; + \; \frac{\cos80}{2\sin40\cos40} \tag3 \\[2ex] &= \frac{2\cos40 \; + \; \cos80}{2\sin40\cos40} \tag4 \\[2ex] &= \frac{\cos40 \; + \; \color{green}{\cos40 \; + \; \cos80}}{\sin80} \tag5 \\[2ex] &= \frac{\cos40 \; + \; \color{green}{2\cos60\cos20}}{\sin80} \tag6 \\[2ex] &= \frac{\cos40 \; + \; \color{green}{\cos20}}{\cos10} \tag7 \\[2ex] &= \frac{2\cos30\cos10}{\cos10} \tag8 \\[2ex] &= 2\cos30 = \sqrt3 \tag9 \\[2ex] \end{align} $$

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  • $\begingroup$ Thanks for the solution! $\endgroup$
    – Aether162
    Jun 18, 2021 at 4:27
  • $\begingroup$ You are welcome! $\endgroup$
    – Saeed
    Jun 18, 2021 at 5:17

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