在三角形ABC中,9a^2+9b^2-19c^2=0.求tanAtanB/[(tanA+tanB)tanC]不要用cot这个符号,没学过

原式=(sinAsinB/cosAcosB)/[(sinA/cosA+sinB/cosB)tanC]
=sinAsinB/((sinAcosB+cosAsinB)tanC)
=sinAsinB/(tanCsin(A+B))
=sinAsinBcosC/(sinCsinC)
=sinAsinB/(sinCsinC)*(a*a+b*b-c*c)/2ab
=sinAsinB/(sinCsinC)*5c*c/9ab
=ab/(c*c)*5c*c/(9ab)
=5/9

因为9a^2+9b^2-19c^2=0，所以a^2+b^2=(19/9)c^2
tanAtanB/[(tanA+tanB)tanC]
(tanA+tanB)/tanAtanB
=1/tanA+1/tanB
=cosA/sinA+cosB/sinB
=(cosAsinB+sinAcosB)/(sinAsinB)
=sin(A+B)/(sinAsinB)=sinC/(sinAsinB)
tanC=sinC/cosC
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

因为9a^2+9b^2-19c^2=0,所以a^2+b^2=(19/9)c^2tanAtanB/[(tanA+tanB)tanC](上下同时除以tanAtanB)=1/(1/tanA+1/tanB)*1/tanC1/tanA+1/tanB=cosA/sinA+cosB/sinB =(cosAsinB+sinAcosB)/(sinAsinB) =sin(A+B)/(sinAsin...