If 2 tan α = 3 tan β, then tan (α−β)=
sin 2β5−cos 2β
2 tan α=3 tan β
Now,
tan(α−β)=tanα−tan β1+tan α tan β=32 tan β−tan β1+(32tan β)tan β=3 tan β−2 tan β2+3 tan2 β=tan β2+3 tan3 β=ssin βcos β2+3sin2 βcos2 β
=sin β cos β2 cos2 β+3sin2 β=sin β cos β3+sin2 β=2 sin β cos β4+2+sin2 β
=sin 2β4+1−cos 2β∴ tan (α−β)=sin 2β5−cos 2β