a+c=b√2,则有
sinA+sinC=√2*sinB,
2*sin[(A+C)/2]*cos[(A-C)/2]=√2*2*sin(B/2)*cos(B/2),
而,(A+C)=(180-B),sin[(A+C)/2]=sin[90-(B/2)]=cos(B/2),则有
2cos[(A-C)/2]=√2*sin(B/2),
而,B/2=90-(A+C)/2,sin(B/2)=sin[90-(A+C)/2]=cos[(A+C)/2],则有
2*cos[(A-C)/2]=√2*cos[(A+C)/2],
2*[cos(A/2)*cos(C/2)+sin(A/2)*sin(C/2)]=√2*[cos(A/2)*cos(C/2)-sin(A/2)*sin(C/2)],
(2+√2)*[sin(A/2)*sin(C/2)]=(√2-2)*[cos(A/2)*cos(C/2)],
[sin(A/2)*sin(C/2)]/[cos(A/2)*cos(C/2)]=(√2-2)/(√2+2)=2√2-3.
即:tan(A/2)*tan(C/2)=2√2-3.