Question

在三角形ABC中,若a+c=根号2乘以b,求tan2分之A乘以tan2分之C

Answer 1

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