Question

帮我解一下：已知∠A，∠B，∠C是△ABC的内角。求证：tanA/2×tanB/2+tanB/2×tanC/2+tanC/2×tanA/2=1

高一的同学，脑筋比较不好使，请讲的简单点，谢谢！

Answer 1

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

tanB/2=tan(π-A-C)/2=tan[π/2-(A+C)/2]=cot(A+C)/2
=(1-tanA/2×tanC/2)/(tanA/2+tanC/2)
因此tanA/2×tanB/2+tanB/2×tanC/2+tanA/2×tanC/2
=tanB/2(tanA/2+tanC/2)+tanA/2tanC/2
=[(1-tanA/2×tanC/2)/(tanA/2+tanC/2)]×(tanA/2+tanC/2)+tanA/2tanC/2
=1-tanA/2×tanC/2+tanA/2×tanC/2
=1