# A curious identity on sums of secants

I was working on proving a variant of Markov's inequality, and in doing so I managed to come across an interesting (conjectured) identity for any $n\in\mathbb{N}$:

$$\sum_{m=0}^{n-1} \sec^2\left(\dfrac{(2m+1)\pi}{4n}\right)=2n^2.$$

I tried to prove this via induction, averaging arguments, trig identities, etc., but to no avail. Are there any suggestions on where this identity may be proven or how I should proceed?

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Have you tried pairing up terms for $m=0$ and $m=n-1$, terms for $m=1$ and $m=n-2$ etc.? –  Dilip Sarwate Oct 23 '12 at 3:30
Yes, I had tried that, but things do not work out as symmetrically as I would have hoped. –  Matt Spencerman Oct 23 '12 at 3:31

I checked your sum and it is indeed correct. There is another one you can do too

$$\sum_{m=0}^{n-1} \sec^2\left(\frac{m\pi}{2n}\right) = \sum_{m=0}^{n-1} \sec^2\left(\frac{2m\pi}{4n}\right) = \frac{1}{3}(n^2+1)$$

Combining the two, you will obtain

$$\sum_{m=0}^{2n-1} \sec^2\left(\frac{m\pi}{4n}\right) = \frac{1}{3}(8n^2+1)$$

More specifically

$$\sum_{m=0}^{n-1} \sec^2\left(\frac{m\pi}{2n}\right) = \frac{1}{3}(2n^2+1)$$

These and others can be done with the methods from this question

Prove that $\sum_{k=1}^{n-1}\tan^{2}\frac{k \pi}{2n} = \frac{(n-1)(2n-1)}{3}$

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