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.