<s>[INST] Hints to prove $\tan^2{\theta}=\tan{A}\tan{B}$, given $\frac{\sin{(\theta + A)}}{\sin{(\theta + B)}} = \sqrt{\frac{\sin{2A}}{\sin{2B}}}$ I need some hints on solving this trigonometry problem.

Problem
If $\dfrac{\sin{(\theta + A)}}{\sin{(\theta + B)}} = \sqrt{\dfrac{\sin{2A}}{\sin{2B}}}$, then prove that $\tan^2{\theta}=\tan{A}\tan{B}$.

I tried to expand the left hand side of the equation, but no clue what to do next. I also tried to use $\sin{2\alpha} = \dfrac{2\tan{\alpha}}{1 + \tan^2{\alpha}}$ for the right hand side of the equation with no result.
I appreciate for any help. Thank you.
 [/INST] $$\dfrac{\sin(\theta+A)}{\sin(\theta+B)}=\cdots=\dfrac{\tan\theta\cos A+\sin A}{\tan\theta\cos B+\sin B}$$  Dividing numerator & the denominator by $\cos\theta$ 
$$\implies\dfrac{(\tan\theta\cos A+\sin A)^2}{(\tan\theta\cos B+\sin B)^2}=\dfrac{2\sin A\cos A}{2\sin B\cos B}$$
$$\iff\dfrac{(\tan\theta+\tan A)^2}{(\tan\theta+\tan B)^2}=\dfrac{\tan A}{\tan  B}$$
Dividing numerator of both sides by $\cos^2A$
and the denominator  of both sides by $\cos^2B$
Now simplify assuming $\tan A\ne\tan B$
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