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.