已知3sin2A+B/2+cos2A−B2=2,(cosA•cosB≠0),则tanAtanB=_.
问题描述:
已知3sin2
+cos2A+B 2
=2,(cosA•cosB≠0),则tanAtanB=______. A−B 2
答
3sin2
+cos2A+B 2
=3×A−B 2
+1−cos(A+B) 2
=1+cos(A−B) 2
=2,4−3cos(A+B)+cos(A−B) 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