Prove $(1+ \tan{A}\tan{2A})\sin{2A} = \tan{2A}$. Prove the following statement
$$ (1+ \tan{A}\tan{2A})\sin{2A} = \tan{2A}. $$
On the left hand side I have put the value of $\tan{2A}$ and have then taken the LCM.
I got $\sin{2A}\cos{2A}$. How do I proceed?
Thanks

\begin{align}
& (1 + \tan A \tan(2A))\sin(2A) \\
= & \left(1 + \frac{\sin A}{\cos A}\frac{\sin(2A)}{\cos(2A)}\right)\sin(2A)\\
= & \left(1 + \frac{\sin A}{\cos A}\frac{2\sin A\cos A} 
{\cos(2A)}\right)\sin(2A) \\
= & \left(1 + \frac{2\sin^2 A}{\cos(2A)}\right)\sin(2A) \\
= & \frac{\cos(2A) + 2\sin^2 A}{\cos(2A)}\sin(2A) \\
= & \frac{1 - 2\sin^2 A + 2\sin^2 A}{\cos(2A)}\sin(2A) \\
= & \frac{\sin(2A)}{\cos(2A)} \\
= & \tan(2A)
\end{align}
