由A+B=-π/4 tan(A+B)=-1 即(tanA+tanB)/(1-tanA*tanB)=-1 tanA+tanB=tanA*tanB-1 (tanA-1)(tanB-1)=tanA*tanB-(tanA+tanB)+1 =tanA*tanB-(tanA*tanB-1)+1 =tanA*tanB-tanA*tanB+1+1 =2
