在三角形ABC,已知(tgA-tgB)/(tgA+tgB)=(c-b)/c,求證:B,A,C成等差數列

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證：左邊=(sinA/cosA-sinB/cosB)/(sinAcosA+sinB/cosB)=(sinAcosB-cosAsinB)/(sinAcosB+cosAsinB)=sin(A-B)/sin(A+B)右邊=(2RsinC-2RcosB/(2RsinC)=(sinC-sinB)/sinC=(sinC-sinB)/sin(A+B)---sin(A+B)=sinC-sinB---sinC=sinC-sinB---sinB=0 ???此題錯誤！