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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
  • @RiverLi thanks, that is interesting. – A-Level Student Jan 25 '21 at 15:03

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