Question

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

Answer 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