(2011•郑州三模)在△ABC中,tanA=12,cosB=31010,则tanC的值是( ) A.-1 B.1 C.3 D.2
问题描述:
(2011•郑州三模)在△ABC中,tanA=
,cosB=1 2
,则tanC的值是( )3
10
10
A. -1
B. 1
C.
3
D. 2
答
sinB=
=
1−cos2B
,tanB=
10
10
=sinB cosB
1 3
tanC=tan(180°-A-B)=-tan(A+B)=-
=-1tanA+tanB 1−tanAtanB
故选A