tan(B/2)tan(C/2)+tan(A/2)tan(B/2)+tan(A/2)tan(C/2)=A+B+C=180

问题描述:

tan(B/2)tan(C/2)+tan(A/2)tan(B/2)+tan(A/2)tan(C/2)=A+B+C=180

tan(b/2)tan(c/2)+tan(c/2)tan(a/2)
=tan(c/2)[tan(a/2)+tan(b/2)]
=tan[90-(a+b)/2]×[tan(a/2)+tan(b/2)]
=cot[(a+b)/2]×[tan(a/2)+tan(b/2)]
=[tan(a/2)+tan(b/2)]/tan(a/2+b/2)
=1-tan(a/2)tan(b/2)
∴tan(a/2)tan(b/2)+tan(b/2)tan(c/2)+tan(c/2)tan(a/2) = tan(a/2)tan(b/2)+1-tan(a/2)tan(b/2) = 1