tanA/tanB=sinAcosB/sinBcosA=a^2/b^2=sin^2A/sin^2B
等式两端消去相同项,得sinBcosB=sinAcosA,即2sinBcosB=2sinAcosA,即sin2A=sin2B,即2A+2B=pai,即A+B=pai/2,故三角形为直角三角形
由sinA=2sinBcosC得a=2b*(a^2+b^2-c^2)/2ab化简得b=c,再由sin^2A=sin^2B+sin^2C得a^2=b^2+c^2,从而三角形为等腰直角三角形