I am interest to prove the equation below : $$ \sum_{k=1}^m \tan^2\left(\frac{k\pi}{2m+1}\right) = m(2m+1) $$ you can understand better the first member of the equation here:
(mark the whole url with your mouse because i dont know why the link isn't blue at all) sorry but i am not familiar writing equations here . i hope to understand. so what's the problem in formula i cannot eliminate the trigonometric functions to prove this series is a polynomial . any idea how to manipulate the formula ?
This answer is a nearly identical copy of falagar's beautiful answer here, all credit goes to him.
For the sake of legibility let $x_k:=\tfrac{k\pi}{2m+1}$ for $k=1,\ldots,m$. By Euler's formula we have the identity $$(\cos x_k+i\sin x_k)^{2m+1}=(-1)^k,$$ for each $k$. The binomial expansion of the left hand side gives us $$\sum_{j=0}^{2m+1}\binom{2m+1}{j}(i\sin x_k)^j(\cos x_k)^{2m+1-j}=(-1)^k.$$ Taking the imaginary parts of both sides shows that the terms with odd $j$ sum to zero: $$\sum_{j=0}^m\binom{2m+1}{2j+1}(-1)^j(\sin x_k)^{2j+1}(\cos x_k)^{2m-2j}=0.$$ Dividing both sides by $(\sin x_k)(\cos x_k)^{2m}$ we find that $$\sum_{j=0}^m\binom{2m+1}{2j+1}(-1)^j(\tan x_k)^{2j}=\sum_{j=0}^m\binom{2m+1}{2j+1}\left(-(\tan x_k)^2\right)^j=0,$$ This means that $-(\tan x_k)^2$ is a root of the polynomial $$\sum_{j=0}^m\binom{2m+1}{2j+1}X^j=0,$$ for $k=1,\ldots,m$, and the values of $-(\tan x_k)^2$ are distinct for distinct values of $k$. Hence these are all roots of the polynomial above, and by Vieta's formulas its roots sum to $$-\frac{\tbinom{2m+1}{2m-1}}{\tbinom{2m+1}{2m+1}}=-\binom{2m+1}{2m-1}=-m(2m+1).$$