I've seen many trigonometric identities but here is one that I encountered for which I did not find a reference.


In case you wonder where this came from, I was investigating certain $q$-series [in this paper](https://arxiv.org/pdf/1203.6736.pdf) on Eulerian polynomials.



> 
> **Question.** Is this true? Can you provide a reference or a proof?
>  $$\sum\_{k=-n}^n\frac{2k+1}{n+k+1}\binom{2n}{n-k}\sec((2k+1)\theta)=\sec((2n+1)\theta).$$
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One trig "survives" a binomial summation: why?
The sum from k=-n to k=n-1 equals zero by pairing off the end terms and by using the sign difference together with the evenness of the function sec.


