Question

在锐角三角形ABC，A,B,C的对边分别为a,b,c，b/a+a/b=6cosC,则tanC/tanA+tanC/tanB=

Answer 1

解：因为b/a+a/b=6cosC，所以cosC=(a^2+b^2）/6ab,而cosC=(a^2+b^2-c^2)/2ab,所以3c^2=2(a^2+b^2)
tanC/tanA+tanC/tanB=c/a(cosA/cosC)+c/b(cosB/cosC)
将cosA和cosB用余弦公式展开，并将上面的两个式子代入即得，答案为4
这里还用了sinA/sinC=a/c，sinB/sinC=b/c

Answer 2

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