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证在△ABC中,tanA/2*tanB/2+tanB/2*tanC/2+tanA/2*tanC/2=1
人气:202 ℃ 时间:2019-10-23 02:59:21
解答
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...
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