
①tanA·cotB=1 ②1<sinA+sinB≤③sin2A+cos2B=1 ④cos2A+cos2B=sin2C
A.①③ B.②④ C.①④ D.②③
①tanA·cotB=1 ②1<sinA+sinB≤③sin2A+cos2B=1 ④cos2A+cos2B=sin2C
A.①③ B.②④ C.①④ D.②③
B
解析:∵=sinC
,∴
.
∴tanAcotB=tanAcot(-A)=tan2A,故①错;
sinA+sinB=sinA+sin(-A)=sinA+cosA=
sin(A+
),
∵0<A<,
∴<A+
<
,
∴1<2sin(A+)≤
,故②对,显然③错.
∵cos2A+cos2B=cos2A+sin2A=1=sin2C,∴④对,故选B.