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](/uploads/image/z/19750574-38-4.jpg?t=%E5%B7%B2%E7%9F%A5sin%28x%2By%29%3D1%2F3%2Csin%28x-y%29%3D1%2F5%2C%E5%88%99tanx%2Ftany)
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
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
若sin(x+y)=1/2,sin(x-y)=1/3,则tanx/tany等于?
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