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

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