How to compute the sum of $\cot^2\left(\frac{\pi}{9}\right)+\cot^2\left(\frac{2\pi}{9}\right)+\cot^2\left(\frac{4\pi}{9}\right)=~?$
The answer is $9$.
I tried to use the formula $\cot (2\theta)=\dfrac{\cot^2\theta-1}{2\cot\theta}$ but it is getting more and more complicated.
Let $S= \cot^2\frac{\pi}{9}+\cot^2\frac{2\pi}{9}+\cot^2\frac{4\pi}{9}$ and note that $\cot \frac{4\pi}9=\tan\frac{\pi}{18}$
$$\begin{align} S&=\frac{1}{\sin^2\frac{\pi}9} -1+\frac{1}{\sin^2\frac{2\pi}9}-1+\frac{1}{\cos^2\frac{\pi}{18}}-1 \\ \\ &=\frac{1}{\sin^2\frac{2\pi}9}+\frac{1+4\sin^2\frac{\pi}{18}}{\sin^2\frac{\pi}{9}}-3\\ \\ &=\frac{1+4\cos^2\frac{\pi}{9}\left(1+4\sin^2\frac{\pi}{18}\right)}{\sin^2\frac{2\pi}{9}}-3\\ \\ &=\frac{7+6\cos\frac{2\pi}{9}-4\cos\frac{\pi}{9}-4\cos\frac{\pi}{9}\cos\frac{2\pi}{9}}{\sin^2\frac{2\pi}{9}}-3\\ \\ &=\frac{6-6\cos\frac{\pi}{9}+6\cos\frac{2\pi}{9}}{\sin^2\frac{2\pi}{9}}-3\\ \\ &=\frac{6\cos\frac{\pi}{9}\left(2\cos\frac{\pi}{9}-1\right)}{\sin^2\frac{2\pi}{9}}-3\\ \\ &=\frac{24\cos\frac{\pi}{9}\left(\sin\frac{2\pi}{9}\sin\frac{\pi}{9}\right)}{\sin^2\frac{2\pi}{9}}-3\\ \\ &=12-3\\ \\ &=9 \end{align}$$
L.H.S. $=\cot^2(20°) +\cot^2(40°) +\cot^2(80°).$
Let $A = 20°$ then $3A = 60°$ hence $\cot(3A)=\cot(60°)= \frac1{\sqrt3}.$
L.H.S $=\cot^2 A +\cot^2 (60°-A) +\cot^2 (60° + A).$
Write in terms of tan and use tan(A-B) formula to simplify.
$=\frac1{\tan^2 A} + \frac1{\tan^2 (60°-A)} +\frac1{\tan^2 (60° + A)}.$
$= {(6tan^6 A} + {45 tan^4A} +9)/(3 tanA - tan^3A)^2.$
Replace $\tan^6A$ with the help of $\tan^2(3A)=3$
$ tan^6 A = 33 tan^4A - 27tan^2A +3.$
Final step is $27\cot^2(3A)=\frac{27}3 = 9.$
