How to prove a trigonometric identity $\tan(A)=\frac{\sin2A}{1+\cos 2A}$ Show that
$$
\tan(A)=\frac{\sin2A}{1+\cos 2A}
$$
I've tried a few methods, and it stumped my teacher.

First, lets develop a couple of identities.
Given that $\sin 2A = 2\sin A\cos A$, and $\cos 2A = \cos^2A - \sin^2 A$ we have
$$\begin{array}{lll}
\tan 2A &=& \frac{\sin 2A}{\cos 2A}\\
&=&\frac{2\sin A\cos A}{\cos^2 A-\sin^2A}\\
&=&\frac{2\sin A\cos A}{\cos^2 A-\sin^2A}\cdot\frac{\frac{1}{\cos^2 A}}{\frac{1}{\cos^2 A}}\\
&=&\frac{2\tan A}{1-\tan^2A}
\end{array}$$
Similarly, we have
$$\begin{array}{lll}
\sec 2A &=& \frac{1}{\cos 2A}\\
&=&\frac{1}{\cos^2 A-\sin^2A}\\
&=&\frac{1}{\cos^2 A-\sin^2A}\cdot\frac{\frac{1}{\cos^2 A}}{\frac{1}{\cos^2 A}}\\
&=&\frac{\sec^2 A}{1-\tan^2A}
\end{array}$$
But sometimes it is just as easy to represent these identities as

$$\begin{array}{lll}
(1-\tan^2 A)\sec 2A &=& \sec^2 A\\
(1-\tan^2 A)\tan 2A &=& 2\tan A
\end{array}$$

Applying these identities to the problem at hand we have
$$\begin{array}{lll}
\frac{\sin 2A}{1+\cos 2A}&=& \frac{\sin 2A}{1+\cos 2A}\cdot\frac{\frac{1}{\cos 2A}}{\frac{1}{\cos 2A}}\\
&=& \frac{\tan 2A}{\sec 2A +1}\\
&=& \frac{(1-\tan^2 A)\tan 2A}{(1-\tan^2 A)(\sec 2A +1)}\\
&=& \frac{(1-\tan^2 A)\tan 2A}{(1-\tan^2 A)\sec 2A +(1-\tan^2 A)}\\
&=& \frac{2\tan A}{\sec^2 A +(1-\tan^2 A)}\\
&=& \frac{2\tan A}{(\tan^2 A+1) +(1-\tan^2 A)}\\
&=& \frac{2\tan A}{2}\\
&=& \tan A\\
\end{array}$$
Lessons learned: As just a quick scan of some of the other answers will indicate, a clever substitution can shorten your workload considerably.
