怎样证明cot(A/2)+cot(B/2)+cot(C/2)=cot(A/2)cot(B/2)cot(C/2),想知道详细过程

Question

怎样证明cot(A/2)+cot(B/2)+cot(C/2)=cot(A/2)cot(B/2)cot(C/2),想知道详细过程

Answer 1

A,B,C应该是指三角形的内角吧.
A+B+C=180°
A+B=180°-C
(A/2)+(B/2)=90°-(C/2)
tan(A/2+B/2)=tan(90°-C/2)
=cot(C/2)
=(tan(A/2)+tan(B/2))/(1-tan(A/2)(tan(B/2))
化简
tan(A/2)+tan(B/2)=cot(C/2)(1-tan(A/2)(tan(B/2))
1/cot(A/2)+1/cot(B/2)=cot(C/2)-cot(C/2)/[cot(A/2)(cot(B/2)]
两边同时乘以cot(A/2)cot(B/2),得
cot(A/2)+cot(B/2)=cot(C/2)cot(A/2)cot(B/2)-cot(C/2)
故
cot(A/2)+cot(B/2)+cot(C/2)=cot(A/2)cot(B/2)cot(C/2)
如仍有疑惑,欢迎追问.祝：