问题 填空题
锐角△ABC中,
b
a
+
a
b
=6cosC
,则
tanC
tanA
+
tanC
tanB
=______.
答案

b
a
+
a
b
=
a2+b2
ab
=6cosC,

由余弦定理得:a2+b2-2abcosC=c2

∴4ab•cosC=c2

则原式=tanC•

sinBcosA+sinAcosB
sinAsinB
=tanC•
sin(A+B)
sinAsinB
=
sin2C
sinAsinBcosC

由正弦定理得:

sin2C
sinAsinB
=
c2
ab

∴上式=

c2
abcosC
=
4abcosC
abcosC
=4.

故答案为:4

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