# Trigonometric idnetity

IF $\sin \alpha = 3 \sin (\alpha+2\beta)$, then the value of $\tan (\alpha+\beta)+2 \tan \beta=$?

ATTEMPT: $\sin \alpha = 3 (\sin (\alpha+\beta) \cos \beta + \cos (\alpha+\beta) \sin \beta)$

Dividing by $\cos\beta \cos(\alpha+\beta)$ $$\frac{\sin \alpha}{\cos \beta \cos(\alpha+\beta)}=3[\tan(\alpha+\beta)+\tan\beta]$$ Putting $\tan(\alpha+\beta)+2\tan \beta= X$ $$\frac{\sin\alpha}{\cos\beta \cos(\alpha+\beta)}=3[X-\tan\beta]$$

$$SinA=3\left(SinA\,Cos2B+CosA\,Sin2B\right)$$ Divide throughout by $CosA$ we get

$$TanA=3\left(TanA\,Cos2B+Sin2B\right)$$ Use $Cos2B$ and $Sin2B$ in terms of $Tan$ and then Simplify

$$\frac{\sin(\alpha+2\beta)}{\sin\alpha}=\frac13$$

Applying Componendo and dividendo, $$\frac{\sin(\alpha+2\beta)-\sin\alpha}{\sin(\alpha+2\beta)+\sin\alpha}=\frac{1-3}{1+3}$$

Applying Prosthaphaeresis Formulas,

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

Can you take it home from here?