Question

【高中数学】已知AB都是锐角,且A+B≠π/2,(1+tanA)(1+tanB)=2,求证A+B=π/4

Answer 1

(1+tanA)(1+tabB)=2

1+tanA+tanB+tabAtabB=2

tanA+tanB+tanAtanB=1

sinA/cosA+sinB/cosB+sinA/cosA*sinB/cosB=1

sinAcosB+cosAsinB+sinAsinB-cosAcosB=0

sin(A+B)-cos(A+B)=0

sin(A+B)=cos(A+B)

A+B≠π/2, cos(A+B)≠0

所以 tan(A+B)=1

A+B=π/4