题目

△ABC中,三边a,b,c,且a+c=2b,则tanA/2*tanC/2=_________

提问时间：2021-01-31

答案

因为a+c=2b 所以sinA+sinC=2sinB
2sin[(A+C)/2]cos[(A-C)/2]=2*2sin(B/2)cos(B/2)
2sin[(180-B)/2]cos[(A-C)/2]=2*2sin(B/2)cos(B/2)
cos(B/2)cos[(A-C)/2]=2sin(B/2)cos(B/2)
cos[(A-C)/2]=2sin(B/2)
tan(A/2)*tan(C/2)=sin(A/2)sin(C/2)/[cos(A/2)cos(C/2)]
=-1/2{cos[(A+C)/2]-cos[(A-C)/2]}/(1/2{cos[(A+C)/2]+cos[(A-C)/2]})
=-{sin(B/2)-cos[(A-C)/2]}/{sin(B/2)+cos[(A-C)/2]}
=-[sin(B/2)-2sin(B/2)]/[sin(B/2)+2sin(B/2)]
=1/3