锐角三角形ABC中b比a+a比b=6cosC,求tanC乘cotA+tanC乘cotB

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锐角三角形ABC中b比a+a比b=6cosC,求tanC乘cotA+tanC乘cotB

b/a+a/b=6cosC
b^2+a^2=6abcosC
cosC=(ab^2+a^2-c^2)/(2ab)
2abcosC=ab^2+a^2-c^2
4abcosC=c^2
sin²C/4=sinAsinBcosC
tanC*cotA+tanC*cotB
=tanC(cotA+cotB)
=sinC/cosC(cosA/sinA+cosB/sinB)
=sinC/cosC(cosAsinB+sinAcosB)/(sinAsinB)
=sinC/cosC*sin(B+A))/(sinAsinB)
=sin²C/(cosCsinAsinB)
=sin²C/(sin²C/4)=4