Question

在三角形ABC内,证明 tan(A/2)+tan(B/2)-cot(C/2)= -tan(A/2)tan(B/2)cot

Answer 1

tan(A/2)+tan(B/2)-cot(C/2)

=tan(A/2)+tan(B/2)-tan(π/2-C/2)

=tan((A+B)/2)(1-tan(A/2)tan(B/2))-tan((A+B)/2)

=tan((A+B)/2)-tan((A+B)/2)tan(A/2)tan(B/2)-tan((A+B)/2)

=-tan((A+B)/2)tan(A/2)tan(B/2)

=-tan(A/2)tan(B/2)tan(π/2-C/2)

=-tan(A/2)tan(B/2)cot(C/2)