1/[tan(A/2)]=tan(π/4+B/2)由此证A+B=π/2
在△ABC中,已知[tan(A/2)+tan(A/2)*tan(B/2)]/(1-tan(B/2)=1,
则有1/[tan(A/2)]=tan(π/4+B/2),由此证A+B=π/2