Show that if $\alpha$ and $\beta$ are acute angles such that :
$$\left[\sin(\alpha-\beta)+\cos(\alpha+2\beta)\sin \beta\right]^2=4 \cos \alpha \cos \beta\sin(\alpha+\beta)$$
then $$\tan \alpha =\tan \beta \left[\dfrac{1}{(\sqrt{2}\cos \beta-1)^2}-1\right]$$
I tried to use componendo dividendo to prove the statement but got nowhere.
I don't get how to simplify or operate on the $\sqrt{2}\cos\beta - 1$ part.
Hint
$$\sin A(\cos B-\sin B\sin2B)+\cos A(\cos2B\sin B-\sin B)$$ $$=\sin A\cos B\cos2B-2\cos A\sin^3B$$
Divide both sides of the given condition by $\cos^2A\cos^B$ to find
$$(\tan A\cos2B-2\sin^2B\tan B)^2=4(\tan A+\tan B)$$
$$\iff((\tan A+\tan B)\cos2B-\tan B)^2=4(\tan A+\tan B)$$
Rearrange to form a quadratic equation in $\tan A+\tan B$
Finally find the roots of the equation