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

问题描述:

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

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

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...