Question

在三角形中 求证tanA/2tanB/2+tanB/2tanC/2+tanC/2tanA/2=1

Answer 1

证明：由于A,B,C为△ABC中三个内角 ,则:

tanA/2*tanB/2+tanB/2*tanC/2+tanC/2*tanA/2

=tanA/2*tanB/2+tanB/2*tan[pi/2-(A+B)/2]+tan[pi/2-(A+B)/2]*tanA/2

=tanA/2*tanB/2+tanB/2*cot[(A+B)/2]+cot[(A+B)/2]*tanA/2

=tanA/2*tanB/2+cot[(A+B)/2]*[tanA/2+tanB/2]

由于:tan[(A+B)/2]=[tanA/2+tanB/2]/[1-tanA/2*tanB/2]

故:tanA/2+tanB/2=tan[(A+B)/2]*[1-tanA/2*tanB/2]

则原式=tanA/2*tanB/2+cot[(A+B)/2]*{tan[(A+B)/2]*[1-tanA/2*tanB/2]}

=tanA/2*tanB/2 + 1 *(1-tanA/2*tanB/2)

=tanA/2*tanB/2+1-tanA/2*tanB/2=1

原命题得证