TITLE: A curious identity on sums of secants
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: 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 $\sum\limits_{k=1}^{n-1}\tan^{2}\frac{k \pi}{2n} = \frac{(n-1)(2n-1)}{3}$
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