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!