三角恒等变换在△ABC中,求证：tanA/2tanB/2+tanB/2tanC/2+tanC/2tanA/2=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
故原命题得证