求证sin(A+B)/cos(A-B)=(tanA+tanB)/(1+tanAtanB)如题,sin(A+B)/cos(A-B)=(tanA+tanB)/(1+tanAtanB)
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![求证sin(A+B)/cos(A-B)=(tanA+tanB)/(1+tanAtanB)如题,sin(A+B)/cos(A-B)=(tanA+tanB)/(1+tanAtanB)](/uploads/image/z/2650592-56-2.jpg?t=%E6%B1%82%E8%AF%81sin%28A%2BB%29%2Fcos%28A-B%29%3D%28tanA%2BtanB%29%2F%281%2BtanAtanB%29%E5%A6%82%E9%A2%98%2Csin%28A%2BB%29%2Fcos%28A-B%29%3D%28tanA%2BtanB%29%2F%281%2BtanAtanB%29)
求证sin(A+B)/cos(A-B)=(tanA+tanB)/(1+tanAtanB)如题,sin(A+B)/cos(A-B)=(tanA+tanB)/(1+tanAtanB)
求证sin(A+B)/cos(A-B)=(tanA+tanB)/(1+tanAtanB)
如题,
sin(A+B)/cos(A-B)=(tanA+tanB)/(1+tanAtanB)
求证sin(A+B)/cos(A-B)=(tanA+tanB)/(1+tanAtanB)如题,sin(A+B)/cos(A-B)=(tanA+tanB)/(1+tanAtanB)
sin(A+B)/cos(A-B)
=(sinAcosB+cosAsinB)/(cosAcosB+sinAsinB)
=[(sinAcosB+cosAsinB)/cosAcosB]/[(cosAcosB+sinAsinB)/cosAcosB]
=(tanA+tanB)/(1+tanAtanB)