The correct option is A ab
Given :
sin(α+β)sin(α−β) = a+ba−bApplying sin(α+β)=sinαcosβ + cosαsinβ, we get⇒ sinαcosβ + cosαsinβsinαcosβ − cosαsinβ = a+ba−b
Applying componendo and dividendo we get:
⇒(sinαcosβ + cosαsinβ)+(sinαcosβ − cosαsinβ)(sinαcosβ + cosαsinβ)−(sinαcosβ − cosαsinβ)= (a+b)+(a−b)(a+b)−(a−b)
⇒2 sinαcosβ2 cosαsinβ = 2a2b⇒ tanαtanβ = ab