Pls solve q25 fast with explanation Share with your friends Share 0 Premsagar Ram answered this Dear student, We have,1111+sinA1+sinB1+sinCsinA+sin2AsinB+sin2BsinC+sin2CApplying R3→ R3- R2=1111+sinA1+sinB1+sinCsinA+sin2A-1-sinAsinB+sin2B-1-sinBsinC+sin2C-1-sinC=1111+sinA1+sinB1+sinC-1-sin2A-1-sin2B-1-sin2C=1111+sinA1+sinB1+sinC-cos2A-cos2B-cos2CApplying C3→ C3- C2 and C2→ C2- C1=1001+sinAsinB-sinAsinC-sinB-cos2Acos2A-cos2Bcos2B-cos2CExpanding along R1, we get∆=sinB-sinAsin2C-sin2B-sinC-sinBsin2B-sin2A=sinB-sinAsinC+sinBsinC-sinB-sinC-sinBsinB+sinAsinB-sinA=sinB-sinAsinC-sinBsinC+sinB-sinB-sinA=sinB-sinAsinC-sinBsinC-sinA=0 either sinB-sinA=0 or sinC-sinB=0 or sinC-sinA=0⇒A=B or B=C or C=A⇒∆ABC is an issosecles triangle. Regards 0 View Full Answer
