If 3 sinβ = sin(2α+ β), then tan(α+ β) is equal to

Q: If 3 sinβ = sin(2α+ β), then tan(α+ β) is equal to

(A) 2 tanβ

(B) 2 tanα

(C) tanα + tanβ

(D) none of these

Sol. $\large \frac{sin(2\alpha + \beta)}{sin\beta} = \frac{3}{1}$

$\large \frac{sin(2\alpha + \beta) + sin\beta}{sin(2\alpha + \beta) – sin\beta} = \frac{3+1}{3-1} = 2 $

$\large \frac{2 sin(\alpha + \beta) cos\alpha}{2sin\alpha cos(\alpha + \beta)} = 2$

⇒ tan(α + β) = 2 tanα

Hence (B) is the correct answer.