tanB(sinA-sinC)=cosC-cosA
tanB=(cosC-cosA)/(sinA-sinC)
cosC-cosA=2cos[(C+A)/2]cos[(C-A)/2]
sinA-sinC=2cos[(A-C)/2]sin[(A+C)/2]
tanB=-cot[(A+C)/2]
tanB=cot[180-(A+C)/2]=cot[90+(180-A-C)/2]=cot(90+B/2)=tan(-B/2)
tanB=2tan(B/2)/[1-(tanB/2)^2]
tan(-B/2)=2tan(B/2)/[1-tan(B/2)^2]
-1=2/[1-tan(B/2)^2] [tan(B/2)]^2-1=2 ,B