证明：∵2tanα=3tanβ∴tanα=3tanβ/2∴tan(α-β)=(tanα-tanβ)/(1+tanαtanβ)=(3tanβ/2-tanβ)/1+(3tanβ/2)^2=tanβ/(2+3tan^2β)=(sinβ/cosβ)/[(2cos^2β+3sin^2β)/cos^2β]=sinβcosβ/(2+sin^2β)=sin...