在锐角△ABC中,角A、B、C的对边分别为a、b、c,若a/b+b/a=6cosC,则tanC/tanA+tanC/tanB的值是_.

问题描述:

在锐角△ABC中,角A、B、C的对边分别为a、b、c,若

a
b
+
b
a
=6cosC,则
tanC
tanA
+
tanC
tanB
的值是______.

a
b
+
b
a
=6cosC,
由余弦定理可得,
a2+b2
ab
=6•
a2+b2c2
2ab

a2+b2
3c2
2

tanC
tanA
+
tanC
tanB
=
cosAsinC
cosCsinA
+
cosBsinC
cosCsinB
=
sinC
cosC
(
cosA
sinA
 +
cosB
sinB
)

=
sinC
cosC
sinBcosA+sinAcosB
sinAsinB
=
sin2C
sinAsinBcosC
=
c2
abcosC

=
c2
ab
2ab
a2+b2c2
=
2c2
3c2
2
c2
=4

故答案为:4