∵A+B=π/4
∴tan(A+B)=(tanA+tanB)/(1-tanAtanB)=1
则tanA+tanB=1-tanAtanB
即(1+tanA)(1+tanB)
=1+tanA+tanB+tanAtanB
=1+1-tanAtanB+tanAtanB
= 2
∵A+B=π/4
∴tan(A+B)=(tanA+tanB)/(1-tanAtanB)=1
则tanA+tanB=1-tanAtanB
即(1+tanA)(1+tanB)
=1+tanA+tanB+tanAtanB
=1+1-tanAtanB+tanAtanB
= 2