已知3sin2A+B/2+cos2A−B2=2,(cosA•cosB≠0),则tanAtanB=_.

问题描述:

已知3sin2

A+B
2
+cos2
A−B
2
=2,(cosA•cosB≠0),则tanAtanB=______.

3sin2

A+B
2
+cos2
A−B
2
=3×
1−cos(A+B)
2
+
1+cos(A−B)
2
=
4−3cos(A+B)+cos(A−B)
2
=2,
∴4-3cos(A+B)+cos(A-B)=4,即3cos(A+B)=cos(A-B),
∴3cosAcosB-3sinAsinB=cosAcosB+sinAsinB,即2cosAcosB=4sinAsinB,
则tanAtanB=
sinAsinB
cosAcosB
=
2
4
=
1
2

故答案为:
1
2