已知tan（α＋β）=1／3,tan2β=-2；求tan2α的值 tan[2(α+β)]=2tan(α+β)/[1-tan²(α+β)]=(2/3)/(1-1/9)=3/4tan2α=tan[2(α+β)-2β]=[tan2(α+β)-tan2β]/[1+tan2(α+β)tan2β]=(3/4+2)/[1+(3/4)(-2)]...