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=tan...