已知sin（x+y）=1,求证：tan（2x+3y）=tany

已知sin（x+y）=1,
求证：tan（2x+3y）=tany
证明：sin（x+y）=1
所以x+y=2k兀+兀/2 K为整数
所以tan(2x+3y)=tan(4k兀+兀+y)=tan(兀+y)=tany

证明，
tan(2x+3y)
= tan(2x+2y + y)
= [tan(2x+2y) + tany] / (1 - tan(2x+2y)*tany)
因为sin(x+y) = 1
所以cos(x+y) = 0
所以sin(2x+2y) = 2 * sin(x+y) * cos(x+y) = 0
所以 tan(2x+2y) = 0
因此tan(2x+3y)
= (0 + tany) / ( 1+ 0)
= tany / 1 = tany