tan(α+π/6)=1/2,tan(β-7/6π)=1/3,求tan(α+β)

tan(α+π/6)=1/2,tan(β-7/6π)=1/3,求tan(α+β)
tan(α+π/6)=1/2,tan(β-7/6π)=1/3,求tan(α+β)

tan(α+π/6)=1/2,tan(β-7/6π)=1/3,求tan(α+β)
tan(β-7/6π)=tan(β-π-π/6)=tan(β-π/6)
tan(α+β)=tan[(α+π/6）+（β-π/6)]
=[tan(α+π/6)+tan(β-7/6π)]/[1-tan(α+π/6)*tan(β-7/6π)]
=[1/2+1/3]/[1-(1/2)*(1/3)]
=1

由tan(β-7/6π)=1/3得tan(β-1/6π)=1/3
故tan(α+β)=(1/2+1/3)/(1-1/6)=1