The indefinite integral of tan(x) is -ln |cos(x)|, so find out how that expression was arrived at and see if this can be adpated to the discrete case.

http://en.wikipedia.org/wiki/Indefinite_sum

http://en.wikipedia.org/wiki/Table_of_integrals#Trigonometric_functions

Or go for an approximate answer: truncate the taylor series and find the indefinite sum of that polynomial. From http://en.wikipedia.org/wiki/Taylor_series tan(x) = x+x^3/3+... and sum(x+x^3/3,x) = (7/12)x^2-(1/2)x+(1/12)x^4-(1/6)x^3

The indefinite integral of tan(x) is -ln |cos(x)|, so find out how that expression was arrived at and see if this can be adpated to the discrete case.

http://en.wikipedia.org/wiki/Indefinite_sum

http://en.wikipedia.org/wiki/Table_of_integrals#Trigonometric_functions

Or go for an approximate answer: truncate the taylor series and find the indefinite sum of that polynomial. From http://en.wikipedia.org/wiki/Taylor_series tan(x) = x+x^3/3+... and sum(x+x^3/3,x) = (7/12)x^2-(1/2)x+(1/12)x^4-(1/6)x^3

The problem with this is tan(x) is periodically discontinuous and the taylor series is only valid for one period.