已知sin(x+y)=1/3,sin(x-y)=1/5,则tanx/tany
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已知sin(x+y)=1/3,sin(x-y)=1/5,则tanx/tany已知sin(x+y)=1/3,sin(x-y)=1/5,则tanx/tany已知sin(x+y)=1/3,sin(x-y)=
已知sin(x+y)=1/3,sin(x-y)=1/5,则tanx/tany
已知sin(x+y)=1/3,sin(x-y)=1/5,则tanx/tany
已知sin(x+y)=1/3,sin(x-y)=1/5,则tanx/tany
sin(x+y)=sinxcosy+cosxsiny=1/3,sin(x-y)=sinxcosy-cosxsiny=1/5,
两式作商得(sinxcosy+cosxsiny)/(sinxcosy-cosxsiny)=5/3,分子分母同时除以cosxcosy得
(tanx+tany)/(tanx-tany)=5/3,所以3tanx+3tany=5tanx-5tany,tanx=4tany,所以tanx/tany=4
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已知sin(x+y)=1/3,sin(x-y)=1/5,则tanx/tany
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