Question

在锐角三角形ABC中，角A.B.C的对边分别为abc,若b/a+a/b=6cosC,则tanC/tanA +tanC/tanB的值为什么

答案是4，，请写过程

Answer 1

b/a+a/b=6cosC
两边乘以ab
b²+a²=6abcosC
由余弦定理
c²=a²+b²-2abcosC=4abcosC
由正弦定理
c/sinC=b/sinB=a/sinA
代入上式得 sin²C=4sinAsinBcosC
tanC/tanA +tanC/tanB
=sinCcosA/sinAcosC+sinCcosB/sinBcosC
=(sinCcosAsinB+sinCcosBsinA)/sinAsinBcosC
=sinC(cosAsinB+cosBsinA)/sinAsinBcosC
=sinCsin(A+B)/sinAsinBcosC
=sin²C/sinAsinBcosC
=4sinAsinBcosC/sinAsinBcosC=4