We have, acosθ+bsinθ=c.....(i) ∵α and β(α=β) satisfy the Eq. (i) ⇒acosα+bsinα=c..... (ii)
and acosβ+bsinβ=c.....(iii)
Now, subtracting Eq. (iii) from Eq. (ii), we get acosα+bsinα−acosβ−bsinβ=0 ⇒a(cosα−cosβ)+b(sinα−sinβ)=0 ⇒a(cosα−cosβ)=−b(sinα−sinβ) ⇒asin2α+β=−b[−cos(2α+β)] ⇒asin2(α+β)=bcos2(α+β) ⇒tan(2α+β)=ab