<s>[INST] Prove that $\cot^2{(\pi/7)} + \cot^2{(2\pi/7)} + \cot^2{(3\pi/7)} = 5$ Prove that $\cot^2{(\pi/7)} + \cot^2{(2\pi/7)} + \cot^2{(3\pi/7)} = 5$ .
I am sure this is derived from using roots of unity and Euler's complex number function, but I am very uncomfortable in these areas so some help would be great. It is evident that $(a + b + c)^2 - 2(ab + ac + bc) = a^2 + b^2 + c^2$ .
So, using a polynomial of degree 3 and the coefficients on the $x^2$ and $x$ terms will get where we need to be.
 [/INST] Using this,
the roots of  $\displaystyle z^3+z^2-3z-1=0\qquad (1)$ are $\displaystyle 2\cos\frac{2\pi}7, 2\cos\frac{4\pi}7, 2\cos\frac{6\pi}7$
If $\displaystyle\cot^2\frac{r\pi}7=u, \cos\frac{2r\pi}7=\frac{1-\tan^2\frac{r\pi}7}{1+\tan^2\frac{r\pi}7}=\frac{\cot^2\frac{r\pi}7-1}{\cot^2\frac{r\pi}7+1}=\frac{u-1}{u+1}$
$\displaystyle\implies\frac{2(u-1)}{u+1}=2\cos\frac{2r\pi}7($ where $r=1,2,3)$ will satisfy $(1)$
$\displaystyle\implies \left(\frac{2(u-1)}{u+1}\right)^3+\left(\frac{2(u-1)}{u+1}\right)^2-3\left(\frac{2(u-1)}{u+1}\right)-1=0$
On simplification, $\displaystyle 7 u^3-35 u^2+21 u-1=0$ whose roots are $\displaystyle\cot^2\frac{r\pi}7($  where $r=1,2,3)$
Now, use Vieta's formulas, to find $\displaystyle \sum \cot^2\frac{r\pi}7=\frac{35}7$
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