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Can anyone help me find a formal reference for the following identity about the summation of squared tangent function:

$$ \sum_{k=1}^m\tan^2\frac{k\pi}{2m+1} = 2m^2+m,\quad m\in\mathbb{N}^+. $$

I have proved it, however, the proof is too long to be included in a paper. So I just want to refer to some books or published articles.

I also found it to be a special case of the following identity,

$$ \sum_{k=1}^{\lfloor\frac{n-1}{2}\rfloor}\tan^2\frac{k\pi}{n} = \frac16(n-1)(-(-1)^n (n + 1) + 2 n - 1),\quad n\in\mathbb{N}^+ $$

which is provided by Wolfram.

Thank you very much!

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I can't give you a reference but check this question out. math.stackexchange.com/questions/2339/… –  Aleks Vlasev Oct 20 '12 at 5:06
    
Thanks Aleks, it is pretty good to find this proof. However I think I'd better find a reference for it. –  albert Oct 20 '12 at 5:15
    
It's similar to Proof 9 in Robin Chapman's list of ways to evaluate $\zeta(2)$. –  Steven Taschuk Jul 31 '13 at 17:08

2 Answers 2

In fact, this type of formula is related to binomial coefficients. I give a proof of the general case I found in my post Tan binomial formulas from a set S and its k-subset

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Jolley, Summation of Series, formula 445 is $$\sum_{k=0}^{n-1}\tan^2\left(\theta+{k\pi\over n}\right)=n^2\cot^2\left({n\pi\over2}+n\theta\right)+n(n-1)$$ Let $\displaystyle\theta={\pi\over2m+1}$, $n=2m+1$ and we almost have your sum; we have twice your sum, since the angles here go from just over zero to just under $\pi$, while in your sum they go from just over zero to just under $\pi/2$, and $\tan^2\theta=\tan^2(\pi-\theta)$.

Jolley's reference is to page 73 of S L Loney, Plane Trigonometry, Cambridge University Press, 1900. This book is best known from its part in Ramanujan's early education.

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