Câu 1. Ta có: sinAcosAcosCsinC=sinA.sinC+cosA.cosCsinC=cos(A−C)sinC
⇒sinCcosB=cos(A−C)sinC
⇔sin2C=cosB.cos(A−C) Mà B=π−(A+C)
⇔sin2C=−cos(A+C).cos(A−C)
⇔sin2C=−(cosA.cosC+sinA.sinC)(cosA.cosC−sinA.sinC)
⇔sin2C−sin2A.sin2C+cos2A.cos2C=0
⇔sin2C(1−sin2A)+cos2A.cos2C=0
⇔sin2C.cos2A+cos2A.cos2C=0
⇔cos2A(sin2C+cos2C)=0
⇔cosA=0
⇒A=π2