Question

If sinB=3sin(2A+B) Prove that 2tanA+tan(A+B)=0

Answer 1

If $\sin B = 3 \sin (2A + B)$ prove that $2 \tan A + \tan (A + B) = 0$

Proof:

$$2 \tan A + \tan (A + B) = 0$$ $$\mathrm{LHS} = 2 \tan A + \tan (A + B) \quad (1)$$ $$\tan A = \frac{\sin A}{\cos A}$$ $$\tan (A + B) = \frac{\sin (A + B)}{\cos (A + B)}$$

Substituting into (2):

$$\mathrm{LHS} = 2 \frac{\sin A}{\cos A} + \frac{\sin (A + B)}{\cos (A + B)} = \frac{2 \sin A \cos (A + B) + \sin (A + B) \cos A}{\cos (A + B) \cos A}$$ $$\mathrm{LHS} = \frac{\sin A \cos (A + B) + \sin A \cos (A + B) + \sin (A + B) \cos A}{\cos (A + B) \cos A}$$

The compound-angle formulae:

$$\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$$

So

$$\sin A \cos (A + B) + \sin (A + B) \cos A = \sin (2 A + B)$$

Substituting into (2):

$$\mathrm{LHS} = \frac{\sin A \cos (A + B) + \sin (2 A + B)}{\cos (A + B) \cos A}$$

Product-to-sum identity:

$$\sin \alpha \cos \beta = \frac{1}{2} (\sin (\alpha + \beta) + \sin (\alpha - \beta))$$

So

$$\sin A \cos (A + B) = \frac{1}{2} \sin (2 A + B) + \frac{1}{2} \sin (- B) = \frac{1}{2} \sin (2 A + B) - \frac{1}{2} \sin B$$

Substituting into (3):

$$\mathrm{LHS} = \frac{\frac{1}{2} \sin (2 A + B) - \frac{1}{2} \sin B + \sin (2 A + B)}{\cos (A + B) \cos A}$$ $$\mathrm{LHS} = \frac{\frac{1}{2} (3 \sin (2 A + B) - \sin B)}{\cos (A + B) \cos A}$$

Given

$$\sin B = 3 \sin (2 A + B)$$ $$3 \sin (2 A + B) - \sin B = 0$$

Hence

$$\mathrm{LHS} = \mathrm{RHS} = 0$$