在三角形ABC中,A=60°,(1+sin2A)/(cos^2B-sin^2B)=-3,求tanC

(cos^2B-sin^2B)=cos(2B),(1+sin2A)/(cos^2B-sin^2B)=(1+√3/2）/cos2B=-3
cos2B=-(2+√3)/6,cosB=√(1+cos2B)/2=√[(4-√3)/12],sinB=√[1-(cosB)^2]
=√[(8+√3)/12],sinC=sin[180°-（A+B）]=sin(A+B)=sinAcosB+cosAsinB
=√[（12-3√3)/48]+√[(8+√3)/48] ,C=arsin[√[（12-3√3)/48]+√[(8+√3)/48],cscC=1/sinC,cotC=√(cscC)^2-1,
tanC=1/cotC.

(1+sin2A)/(cos^2B-sin^2B)=-3
cos2B=－（2+√3）/6.∠B≈64.231447612°.∠C≈55.76852388°
tanC≈1.46971789.