263Let a+b=θ and a-b=α
So tana=λtanb
So (λ+1/λ-1)sinα=[(tana + tanb)/(tana-tanb)]*sin(a-b)
=sin(a+b)/sin(a-b)*sin(a-b)
=sinθ
If angle θ is divided into two parts such that the tangent of one part is λ times the tangent of the other, and α is their difference,show that sinθ = (λ+1/λ-1)sinα
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