化简 tanA/2tanB/2+tanB/2tanC/2+tanC/2tanA/2

用到两角和正切公式的变形
tan(M+N)=(tanM+tanN)/(1-tanMtanN)
∴ tanM+tanN=tan(M+N)*(1-tanMtanN)
∴ tanB/2tanA/2+tanB/2tanC/2+tanC/2tanA/2
= tan(B/2)[tan(A/2)+tan(C/2)]+tan(C/2)*tan(A/2)
=tan(B/2)*tan(A/2+C/2)*[1-tan(A/2)tan(C/2)]+tan(C/2)tan(A/2)
=tan(B/2)*tan(π/2-B/2)*[1-tan(A/2)tan(C/2)]+tan(C/2)tan(A/2)
=[sin(B/2)/cos(B/2)]*[sin(π/2-B/2)/cos(π/2-B/2)]*[1-tan(A/2)tan(C/2)]+tan(C/2)tan(A/2)
=[sin(B/2)/cos(B/2)]*[cos(B/2)/sin(B/2)]*[1-tan(A/2)tan(C/2)]+tan(C/2)tan(A/2)
=1*[1-tan(A/2)tan(C/2)]+tan(C/2)tan(A/2)
=1-tan(C/2)tan(A/2)+tan(C/2)tan(A/2)
=1

化简 tanA/2tanB/2+tanB/2tanC/2

tanB/2=tan(π-A-C)/2=tan[π/2-(A+C)/2]=cot(A+C)/2
=(1-tanA/2*tanC/2)/(tanA/2+tanC/2)
tanA/2tanB/2+tanB/2tanC/2+tanA/2tanC/2
=tanB/2(tanA/2+tanC/2)+tanA/2tanC/2
=[(1-tanA/2*tanC/2)/(tanA/2+tanC/2)]*(tanA/2+tanC/2)+tanA/2tanC/2
=1-tanA/2tanC/2+tanA/2tanC/2
=1

应该有条件“A、B、C为三角形内角”吧。
如果A+B+C=π，即A、B、C为三角形内角，故：A/2+B/2+C/2=π/2
则：tan(A/2+B/2)=cot(C/2)=1/ tan(C/2)，tan(A/2+C/2)=cot(B/2)=1/ tan(B/2) ，tan(B/2+C/2)=cot(A/2)=1/ tan(A/2)
又：tan(A/2+B/2)=[tan(A...