sin(x+y)=1,则tan(2x+y)+tany=
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sin(x+y)=1,则tan(2x+y)+tany=sin(x+y)=1,则tan(2x+y)+tany=sin(x+y)=1,则tan(2x+y)+tany=sin(x+y)=1那么：x+y=2k
sin(x+y)=1,则tan(2x+y)+tany=
sin(x+y)=1,则tan(2x+y)+tany=
sin(x+y)=1,则tan(2x+y)+tany=
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