Question

已知ABC是三角形的内角,求证tanA/2*tanB/2+tanB/2*tanC/2+tanC/2*tanA/2=1

Answer 1

tanA/2×tanB/2+tanB/2×tanC/2+tanC/2×tanA/2

=tanA/2×tanB/2+tanC/2×(tanA/2+tanB/2)

=tanA/2×tanB/2+tan[90-(A+B)/2]×(tanA/2+tanB/2)

=tanA/2×tanB/2+cot(A/2+B/2)×(tanA/2+tanB/2)

=tanA/2×tanB/2+(tanA/2+tanB/2)/tan(A/2+B/2)

=tanA/2×tanB/2+1-tanA/2×tanB/2

=1