<s>[INST] Trigonometric proof If $4 \tan(\alpha - \beta) = 3 \tan \alpha $, then prove that
$$\tan \beta = \frac{\sin(2 \alpha)}{7 + \cos(2 \alpha)}$$
This is not homework and I've tried everything so I would just like a straight answer thank you in advance. 
 [/INST] Expanding out the left side of the first equation,
$$ \frac{4 \tan \alpha - 4 \tan \beta}{1 + \tan \alpha \tan \beta} = 3 \tan \alpha$$
Thus
$$ \tan \beta = \frac{\tan \alpha}{4 + 3 \tan^2 \alpha}$$
Writing $\tan \alpha = \dfrac{\sin \alpha}{\cos \alpha}$, this becomes
$$ \tan \beta = \frac{\sin \alpha \cos \alpha}{4 \cos^2 \alpha + 3 \sin^2 \alpha}$$
Now use $\sin (2\alpha) = 2 \sin \alpha \cos \alpha$, $\cos^2 \alpha = \dfrac{1+\cos (2\alpha) }{2}$ and $\sin^2 \alpha = \dfrac{1-\cos (2\alpha) }{2}$.
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