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.
 A: $$\dfrac{\sin(\theta+A)}{\sin(\theta+B)}=\cdots=\dfrac{\tan\theta\cos A+\sin A}{\tan\theta\cos B+\sin B}$$  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 $\tan A\ne\tan B$
A: $$\begin{align*}
\frac{\sin{(\theta + A)}}{\sin{(\theta + B)}} &= \sqrt{\frac{\sin{2A}}{\sin{2B}}}\\
\frac{\sin \theta\cos A + \cos\theta\sin A}{\sin \theta\cos B + \cos\theta\sin B} &= \sqrt\frac{\sin A \cos A}{\sin B \cos B}\\
\frac{\tan \theta\cos A + \sin A}{\tan \theta\cos B + \sin B} &= \sqrt\frac{\sin A \cos A}{\sin B \cos B}\\
(\tan \theta\cos A + \sin A)\sqrt{\sin B \cos B}&=(\tan \theta\cos B + \sin B)\sqrt{\sin A\cos A}\\
\tan\theta(\cos A\sqrt{\sin B \cos B} -\cos B\sqrt{\sin A\cos A}) &= -\sin A\sqrt{\sin B \cos B} + \sin B \sqrt{\sin A\cos A}\\
\tan\theta\sqrt{\cos A\cos B}(\sqrt{\cos A\sin B}-\sqrt{\cos B\sin A}) &= \sqrt{\sin A \sin B}(-\sqrt{\sin A \cos B} + \sqrt{\sin B\cos A})\\
\tan\theta &= \sqrt{\frac{\sin A \sin B}{\cos A\cos B}}\\
\tan^2\theta &= \tan A \tan B
\end{align*}$$
Assuming $\sqrt{\cos A\sin B}-\sqrt{\cos B\sin A} \ne 0$.
