已知cos(α-β)=12/13,cos(α+β)=-1/13,求tanαtanβ的值?

问题描述:

已知cos(α-β)=12/13,cos(α+β)=-1/13,求tanαtanβ的值?

cos(A-B)= cosAcosB+sinAsinB =12/13,.1cos(A+B) = cosAcosB-sinAsinB=-1/13.21式+2式得2cosAcosB=11/13cosAcosB=11/261式-2式得2sinAsinB=1sinAsinB=1/2tanAtanB=sinAsinB/cosAcosB=(1/2)/(11/26)=1/2*26/11=13/11...