Question

производная tgx + ctgx

Answer 1

y' = tg' x + ctg' x = 1/(cos^2 (x) ) + (- 1/(sin^2 (x) ) = 1/(cos^2 (x) ) - 1/(sin^2 (x) ) = 1/(sin^2 (x) * cos^2 (x) ) = [(1-cos(2x))/2 ]*[(1+cos(2x))/2] = (1-cos^2 (2x) )/4