Prove $\sin{2\theta} = \frac{2 \tan\theta}{1+\tan^{2}\theta}$ How to prove: $$\sin{2\theta} = \frac{2 \tan\theta}{1+\tan^{2}\theta}$$ Help please. Don't know where to start.
 A: Set $a=b=\theta $ in the identity
$$\begin{equation*}
\sin (a+b)=\sin a\cdot \cos b+\cos a\cdot \sin b
\end{equation*}$$
to get this one
$$\begin{equation*}
\sin 2\theta =2\sin \theta \cdot \cos \theta .
\end{equation*}$$
Then divide the RHS by $\sin ^{2}\theta +\cos ^{2}\theta =1$ and afterwards both
numerator and denominator by $\cos ^{2}\theta \neq 0$
$$\begin{equation*}
\sin 2\theta =\dfrac{2\sin \theta \cdot \cos \theta }{\sin ^{2}\theta +\cos
^{2}\theta }=\dfrac{2\dfrac{\sin \theta \cdot \cos \theta }{\cos ^{2}\theta }}{
\dfrac{\sin ^{2}\theta +\cos ^{2}\theta }{\cos ^{2}\theta }},
\end{equation*}$$
and simplify.
A: Use the following facts: 


*

*$\sin(A+B)= \sin{A} \cdot \cos{B} + \cos{A} \cdot \sin{B}$.

*$\displaystyle\frac{2 \tan\theta}{1+\tan^{2}\theta} = 2 \cdot \frac{\sin\theta}{\cos\theta} \cdot \cos^{2}\theta = 2 \cdot \sin\theta \cdot \cos\theta$
A: $$\sin2\theta$$
$$2\cdot\sin\theta\cdot \cos\theta$$
multiply and divide by $\cos\theta$
$$ 2\cdot \dfrac {\sin\theta}{\cos\theta}\cdot \cos^2\theta$$
$$2\cdot\tan\theta\cdot\cos^2\theta$$
$$\dfrac{2\cdot\tan\theta}{\sec^2\theta}$$
$$\dfrac{2\cdot\tan\theta}{1+\tan^2\theta}$$
