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
3sin2A+B2+cos2A−B2=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ta...