在三角形ABC中,若(tanB-tanC)/(tanB+tanC)=(a-c)/a求证：2B=A+C

Question

在三角形ABC中,若(tanB-tanC)/(tanB+tanC)=(a-c)/a 求证：2B=A+C

Answer 1

(tanB-tanC)/(tanB+tanC)=(a-c)/a
(tanB+tanC-2tanC)/(tanB+tanC)=(a-c)/a
1-2tanC/(tanB+tanC)=1-c/a
2tanC/(tanB+tanC)=c/a=sinC/sinA
2/(cosCtanB+sinC)=1/sinA
2sinA=(sinBcosC+sinCcosB)/cosB
=sin(B+C)/cosB
=sinA/cosB
2sinAcosB=sinA
sinA（2cosB-1）=0
sinA≠0 所以2cosB-1=0得cosB=1/2
得B=60°
A+C=120°=2B
2B=A+C
证毕!