# A complex problem on trigonometry

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.

• math.stackexchange.com/questions/3216316/… – lab bhattacharjee May 7 '19 at 8:29
• Why are you posting the same question twice? (Just want to attract attention?) – YuiTo Cheng May 7 '19 at 8:31
• Please help me fjnd solution to this problem. It was not answered before so i reposted it – Rituraj Tripathy May 7 '19 at 8:31
• Not receiving answers is not a reason for reposting anyway. – YuiTo Cheng May 7 '19 at 8:32
• From What should I do if no one answers my question? : Edit your question to provide status and progress updates. Document your own continued efforts to answer your question. This will naturally bump your question to the homepage and get more people interested in it. – Martin R May 7 '19 at 8:34

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