已知 cos(α+β)=m,tanαtanβ=2\cos \left( {\alpha + \beta }\right) = m,\tan \alpha \tan \beta = 2cos(α+β)=m,tanαtanβ=2 ，则 cos(α−β)=\cos \left( {\alpha - \beta }\right) =cos(α−β)= （ ）A. −3m- {3m}−3mB. −m3- \frac{m}{3}−3mC. m3\frac{m}{3}3mD. 3m{3m}3m