sin (x+y) =1
那么：x+y = 2kπ +π/2
tan(2x+y)+tany = tan [(x +y)+x] + tany
= tan( 2kπ +π/2 +x) +tan y
= tan (π/2 +x) + tan y
= tan(π/2 +x) + tan(2kπ +π/2 -x) (x+y = 2kπ +π/2)
= - cotx + cotx
=0