证明三角形3内角tan(A/2)*tan(B/2)+tan(B/2)*tan(C/2)+tan(C/2)*tan(A/2)=1

问题描述:

证明三角形3内角tan(A/2)*tan(B/2)+tan(B/2)*tan(C/2)+tan(C/2)*tan(A/2)=1

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)(两角和公式)
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