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
 A: $$\text{LHS}= \\ \tan 2A  \left\{{1\over \tan2A} +\tan A\right\}\sin 2A \\ = \tan2A \left\{\frac{\cos 2A}{\sin 2A} + \frac{\sin A}{\cos A}\right\}\sin 2A \\= \tan2A \left\{\frac{\cos 2A\cos A + \sin2A \sin A}{\cos A}\right\} \\= \tan 2A \left\{\frac{\cos(2A -A)}{\cos A}\right\} \\ =\text{RHS}$$
A: \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}
A: Notice, $$LHS=(1+\tan A\tan2A)\sin 2A$$  $$=\left(1+\tan A\frac{2\tan A}{1-\tan ^2A}\right)\frac{2\tan A}{1+\tan ^2A}$$
$$=\left(\frac{1-\tan^2A+2\tan^2 A}{1-\tan ^2A}\right)\frac{2\tan A}{1+\tan ^2A}$$
$$=\left(\frac{1+\tan^2A}{1-\tan ^2A}\right)\frac{2\tan A}{1+\tan ^2A}$$
$$=\frac{2\tan A}{1-\tan ^2A}$$ $$=\tan 2A=RHS$$
A: So this is equivalent to $$\sin 2A=\tan2A \left(1-\sin2A \tan A\right)$$
And $$1-\sin 2A\tan A=1-2\sin^2A=\cos 2A$$ so that everything works out (using $\sin 2A=2\sin A\cos A$ and $\cos 2A=1-2\sin^2A$)
A: Noting that
$$\cot \theta\tan\theta=1$$
We have
$$(1+\tan A\tan 2A)\sin 2A = (1+\tan A\tan 2A)\sin 2A\cot 2A\tan 2A$$
Now just show that 
$$(1+\tan A\tan 2A)\sin 2A\cot 2A = 1$$
:)
