证在△ABC中,tanA/2*tanB/2+tanB/2*tanC/2+tanA/2*tanC/2=1
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/15 15:31:38
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.
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.
证在△ABC中,tanA/2*tanB/2+tanB/2*tanC/2+tanA/2*tanC/2=1
在锐角三角形ABC中,求证tanA/2tanB/2+tanB/2tanC/2+tanC/2tanA/2=1
三角恒等变换在△ABC中,求证:tanA/2tanB/2+tanB/2tanC/2+tanC/2tanA/2=1
在△ABC中,已知tanA:tanB:tanC=1:2:3,求tan(B-A)
高一三角函数练习题在△ABC中,求证tanA/2×tanB/2 + tanB/2×tanC/2 + tanC/2× ta
在斜三角形ABC中tanC/tanA+tanC/tanB=1,则(a^2+b^2)/c^2
在三角形ABC中,已知tanA=2,tanB=1/3,求tanC
在△ABC中,若tanA:tanB:tanC=1:2:3,则∠A=______.
高三数学题寻解~在三角形ABC中,化简tanA/2tanB/2+tanC/2tanB/2+tanC/2tanA/2
已知任意三角形ABC,求证:tanA/2tanB/2+tanB/2tanC/2+tanC/2tanA/2=1(求步骤说明
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
在△ABC中,若tanA(tanB-tanC)=tanBtanC,则(sinA/sinC)^2+(sinB/sinC)^