问题描述: 已知ABC是三角形的内角,求证tanA/2*tanB/2+tanB/2*tanC/2+tanC/2*tanA/2=1天哪……谁会啊…… 1个回答 分类:数学 2014-10-19 问题解答: 我来补答 tanA/2×tanB/2+tanB/2×tanC/2+tanC/2×tanA/2 =tanA/2×tanB/2+tanC/2×(tanA/2+tanB/2) =tanA/2×tanB/2+tan[90-(A+B)/2]×(tanA/2+tanB/2) =tanA/2×tanB/2+cot(A/2+B/2)×(tanA/2+tanB/2) =tanA/2×tanB/2+(tanA/2+tanB/2)/tan(A/2+B/2) =tanA/2×tanB/2+1-tanA/2×tanB/2 =1 展开全文阅读