在三角形ABC中,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

答案是5／9
1，切割化弦
2，约分后用正弦定理
3，化简后再用余弦
代如已知条件得
tanAtanB/(tanA+tanB)tanC =tanatanb/(tana+tanb)[-tan(a+b)]
=-[tanatanb/(tana+tanb)]*(1-tanatanb)/tanatanb
=tanatanb/(tanatanb-1)
=1+1/tanatanb