证在△ABC中,tanA/2*tanB/2+tanB/2*tanC/2+tanA/2*tanC/2=1

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证在△ABC中,tanA/2*tanB/2+tanB/2*tanC/2+tanA/2*tanC/2=1

证在△ABC中,tanA/2*tanB/2+tanB/2*tanC/2+tanA/2*tanC/2=1
证在△ABC中,tanA/2*tanB/2+tanB/2*tanC/2+tanA/2*tanC/2=1

证在△ABC中,tanA/2*tanB/2+tanB/2*tanC/2+tanA/2*tanC/2=1
A/2+B/2+C/2=90°
A/2=90°-(B/2+C/2)
tanA/2 = tan(90°-(B/2+C/2))
= cot(B/2+C/2)=1/tan(B/2+C/2)
=(1-tanB/2tanC/2)/(tanB/2+tanC/2)
所以tanA/2*tanB/2+tanB/2*tanC/2+tanA/2*tanC/2
= tanA/2 (tanB/2+tanC/2) +tanB/2*tanC/2
= (1-tanB/2tanC/2)/(tanB/2+tanC/2)*(tanB/2+tanC/2) +tanB/2*tanC/2
= 1-tanB/2tanC/2+tanB/2*tanC/2=1,
所以结论成立.