已知 sin(A+B)=1\2,sin(A-B)=1\3,求tan(A+B)-tanA-tanB\tanB*tanB*tan(A+B).
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已知 sin(A+B)=1\2,sin(A-B)=1\3,求tan(A+B)-tanA-tanB\tanB*tanB*tan(A+B).
已知 sin(A+B)=1\2,sin(A-B)=1\3,求tan(A+B)-tanA-tanB\tanB*tanB*tan(A+B).
已知 sin(A+B)=1\2,sin(A-B)=1\3,求tan(A+B)-tanA-tanB\tanB*tanB*tan(A+B).
sin(A+B)=1/2,sin(A-B)=1/3,
sinAcosB+cosAcosB=1/2 (1)
sinAcosB-cosAcosB=1/3 (2)
(1)+(2):
2sinAcosB=5/6
(1)-(2):
2cosAsinB=1/6
【tan(A+B)-tanA-tanB]/[tanB*tanB*tan(A+B)]
=[(tanA+tanB)/(1-tanAtanB)-tanA-tanB]/[tanB*tanB*(tanA+tanB)/(1-tanAtanB)]
=[(tanA+tanB)-(1-tanAtanB)(tanA+tanB)]/[tanB*tanB*(tanA+tanB)]
=(tanA+tanB)tanAtanB/【tanB*tanB*(tanA+tanB)】
=1
后面在有定义得情况下是恒等式,不用条件的
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