<s>[INST] 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.
 [/INST] 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.
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