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《中学生数学》2005年10月上P39吴应邦同学提出这样一个三角命题:在任意△ABC中,若a、b、c分别为角A、B、C的对边,S为△ABC的面积,则tanA/2·tanB/2·tanC/2=4S/(a+b+c)2.我觉得原证法多次运用了和差化积公式和正弦定理,有些复杂．通过对该题的研究,我得出另一简证,供大家参考．证明tanA/2·tanB/2·tanC/2 =sinA/1+cosA·sinB/1+cosB·1-cosC/sinC =sinA·2bc/2bc+b2+c2-a2·sinB·2ac/2ac+a2+c2-b2.
“Mathematical Studies for Middle School Students” in October 2005, P39 Wu Yingbang proposed a triangular proposition: In any ABC, if a, b, and c are the opposite sides of angles A, B, and C, respectively, and S is the area of ABC, then tanA / 2 · tanB / 2 · tan C / 2 = 4S / (a + b + c) 2. I think that the original proof method uses the sum and difference product formula and the sine theorem several times, which is somewhat complicated. Through the study of this question, I came to another for your reference. Prove tanA/2·tanB/2·tanC/2=sinA/1+cosA·sinB/1+cosB·1-cosC/sinC=sinA·2bc/2bc+b2+c2-a2·sinB·2ac/2ac+a2 +c2-b2.