in triangle ABC, prove that 
b-c/a = tan B/2 - tan C/2 / tan B/2 + tan C/2  

Hi, Using sine rule asinA=bsinB=csinC= k b-ca= sinB-sinCsinANow RHS, tanB2-tanC2tanB2+tanC2= sinB2cosB2-sinC2cosC2sinB2cosB2+sinC2cosC2= sinB2cosC2-cosB2sinC2sinB2cosC2+cosB2sinC2=sinB-C2sinB+C2=sinB-C2sinB+C2×2cosB+C22cosB+C2= sinB-sinCsinB+C=  sinB-sinCsin180-A= sinB-sinCsinA
Hence RHS = LHS

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