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The formulae that can be used to evaluate series of sines and cosines of angles in arithmetic progressions are well known: $$\sum_{k=0}^{n-1}\cos (a+k d) =\frac{\sin( \frac{nd}{2})}{\sin ( \frac{d}{2} )} \cos \biggl( \frac{ 2 a + (n-1)d}{2}\biggr)$$

$$\sum_{k=0}^{n-1}\sin (a+kd) =\frac{\sin( \frac{nd}{2})}{\sin ( \frac{d}{2} )} \sin\biggl( \frac{2 a + (n-1)d}{2}\biggr)$$ and these can be derived quite easily through use of Euler's relation and De Moivre's theorem or using the product to sum formulae to form a telescoping series.

Question. In the case of $\tan$ however, what closed form formula is there for $$\sum_{k=0}^{n-1}\tan (a+kd)$$ and how would we derive it? I know of no product to sum identities or other similar ones that could usefully be applied here.

Thanks for your help.

A-Level Student
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    for what it's worth, WolframAlpha is unable to give a nice result; the best it does is this. there probably isn't much of a nicer result; the feeling is that $\cos x$ could be written as $(e^{ix}-e^{-ix})/2$ and similarly for $\sin x$, so they easily led to geometric sequences, but the complex exponential form of $\tan x$ is comparatively complicated. – ho boon suan Jan 20 '21 at 01:19
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    If $d = \frac{\pi}{n}$, there is closed form. See: https://math.stackexchange.com/questions/346368/sum-of-tangent-functions-where-arguments-are-in-specific-arithmetic-series – River Li Jan 25 '21 at 14:59

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As @hoboonsuan showed, the formulae are more than complicated (in fact, I presume that we could not avoid the q-polygamma functions).

However, if you want an approximation, since we know that $$\int\tan (a+k\,d)\,dk=-\frac 1d \log (\cos (a+k\, d ))$$ we could, at least, use the simplest form of Euler-MacLaurin summation.

The formula is quite long if $a\neq 0$, but for $a=0$, using $S=\sec (d n)$, $T=\tan(dn)$, i will give $$\frac 1d \,\log(S)+\frac 1 2 T+\sum_{n=0}^3 \alpha_n\, d^{2n+1}$$ the coefficients being $$\large\left( \begin{array}{cc} n & \alpha_n \\ 0 & \frac{S^2-1}{12} \\ 1 & \frac{-S^4-2 S^2 T^2+1}{360} \\ 2 & \frac{2 S^6+11 S^4 T^2+2 S^2 T^4-2}{3780} \\ 3 & \frac{-17 S^8-180 S^6 T^2-114 S^4 T^4-4 S^2 T^6+17}{75600} \\ \end{array} \right)$$

Trying for $d=10^{-3}$, varying the upper bound by steps of $10$ the results are the same for $25$ significant figures up to $n=1410$. For $n=1560$ (where the angle is very close to $\frac \pi 2$, the sum is something like $4575.59\cdots$ and the absolute error is $3.37\times 10^{-10}$ (this is not too much !).

Claude Leibovici
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