在三角形ABC中,9a^2+9b^2-19c^2=0.求tanAtanB/[(tanA+tanB)tanC]
2个回答

因为9a^2+9b^2-19c^2=0,所以a^2+b^2=(19/9)c^2

tanAtanB/[(tanA+tanB)tanC]

=cotC/(cotA+cotB)

cotA+cotB=cosA/sinA+cosB/sinB

=(cosAsinB+sinAcosB)/(sinAsinB)

=sin(A+B)/(sinAsinB)=sinC/(sinAsinB)

cotC=cosC/sinC

所以原式=cosC*sinA*sinB/(sinC)^2

由正弦定理,sinA*sinB/(sinC)^2=sinA/sinC*sinB/sinC=ab/c^2

由余弦定理,cosC=(a^2+b^2-c^2)/2ab=5(c^2)/9ab

所以原式=5(c^2)/9ab*ab/c^2=5/9