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……

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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……

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