Sin β = sin (α+β-α) = sin (α+β) cos α-sin α cos (α+β). It is known that sinβ=sinαcos(α+β), and the two formulas are simultaneous.

2sinαcos(α+β)=sin(α+β)cosα, finishing, and getting 2tanα=tan(α+β).

Tan (α+β) = (tan α+tan β)/(1-tan α tan β), and 2tanα=tan(α+β), finishing, there are

tanβ=tanα/( 1+2tan^2α)= 1/( 1/tanα+2tanα)，

By using the basic inequality, it is obtained that tanβ≤ 1/2√2=√2/4 if and only if.

1/tanα=2tanα, that is, when tanα=√2/2, take the equal sign.

At this time, tan(α+β)=2tanα=√2.