Question

在锐角三角形ABC中角A、B、C的对边分别为a、b、c，若b/a+a/b=6cosC,则tanC/tanA+tanC/tanB的值是（ ）。

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
希望能解决您的问题。

Answer 2

sinB/sinA+sinA/sinB=6cosC
sin(A+C)/sinA+sin(B+C)/sinB=6cosC
(sinAcosC+cosAsinC)/sinA+(sinBcosC+cosBsinC)/sinB=6cosC
(cosC+sinC/tanA)+(cosC+sinC/tanB)=6cosC
(1+tanC/tanA)+(1+tanC/tanB)=6
tanC/tanA+tanC/tanB=4