在锐角△ABC中,ABC的对边分别为abc,sinB/sinA+sinA/sinB=-6cos(A+B),则tanC/tanA+tanC/tanB= 求详解
问题描述:
在锐角△ABC中,ABC的对边分别为abc,sinB/sinA+sinA/sinB=-6cos(A+B),则tanC/tanA+tanC/tanB=
求详解
答
化简sinB/sinA+sinA/sinB=-6cos(A+B)为[(sinB)^2+(sinA)^2]=6cosCsinAsinB,由正弦定理可得
a^2+b^2=6abcosC
而所求的tanC/tanA+tanC/tanB=sinCcosA/cosCsinA+sinCcosB/cosCsinB
==sinCsin(A+B)/cosCsinAsinB=(sinC)^2/cosCsinAsinB=c^2/abcosC
由余弦定理,得c^2=a2+b^2-2abcosC=4abcosC
从而所求的式子的值为4