解三角形函数的
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解三角形函数的
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/a+a/b=(a^2+b^2)/ab=6cosC
根据余弦定理:a^2+b^2-2abcosC=c^2
=>4ab cosC=c^2
tanC/tanA+tanC/tanB
=tanC[(sinBcosA+sinAcosB)/sinAsinB]
=tanC*sin(A+B)/(sinAsinB)
=tanC*sinC/(sinA*sinB)
=(sinC)^2/(sinA*sinB*cosC)
由正弦定理,(sinC)^2/(sinAsinB)=c^2/(ab)
所以上式 = c^2/(ab cosC)
又 4ab cosC=c^2
所以上式 = 4
根据余弦定理:a^2+b^2-2abcosC=c^2
=>4ab cosC=c^2
tanC/tanA+tanC/tanB
=tanC[(sinBcosA+sinAcosB)/sinAsinB]
=tanC*sin(A+B)/(sinAsinB)
=tanC*sinC/(sinA*sinB)
=(sinC)^2/(sinA*sinB*cosC)
由正弦定理,(sinC)^2/(sinAsinB)=c^2/(ab)
所以上式 = c^2/(ab cosC)
又 4ab cosC=c^2
所以上式 = 4