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.
 A: Proof without words: $\tan(A)=\dfrac{\color{red}{\sin(2A)}}{\color{blue}{1}+\color{green}{\cos(2A)}}$
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A: We need to prove that:
$$\frac{\sin(2A)}{1+\cos(2A)}=\tan(A)$$
Let's do LHS-RHS to prove it. I will try to make the left side equal the right side.
$$\frac{\sin(2A)}{1+\cos(2A)}$$
Using double angle identites for both sine and cosine:
$$\frac{2\sin(A)\cos(A)}{1+2\cos^2(A)-1}$$
How nice. The $1$ and $-1$ in the denominator cancel out.
$$\frac{2\sin(A)\cos(A)}{2\cos^2(A)}$$
Cancelling out the $\cos(A)$ in the numerator and the denominator yields:
$$\frac{2\sin(A)}{2\cos(A)}$$
We can also cancel out the $2$ in the numerator and the denominator.
$$\frac{\sin(A)}{\cos(A)}$$
$$=\tan(A)$$
$$\text{LHS=RHS}$$
$$\displaystyle \boxed{\therefore \dfrac{\sin(2A)}{1+\cos(2A)}=\tan(A)}$$
A: $$\sin 2A = 2 \sin A \cos A$$
$$\cos 2A = 2 \cos^2A - 1$$
Substitute these identities and you will get $\tan A$. 
A: 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.
A: The given equality is false. Set $A = \pi/2$. (Note: this applied to an earlier version of the problem).
Perhaps what you meant was
$$ \tan \frac{A}{2} = \frac{\sin A}{1 + \cos A}$$
or
$$ \tan A = \frac{\sin 2A}{1 + \cos 2A}$$
which is true, by using the half/double angle formulas.
$$\frac{\sin A}{1 + \cos A} = \frac{ 2 \sin A/2 \cos A/2}{2 \cos^2 A/2} = \tan A/2$$
