How to prove: $$\sin{2\theta} = \frac{2 \tan\theta}{1+\tan^{2}\theta}$$ Help please. Don't know where to start.
[Math] Prove $\sin{2\theta} = \frac{2 \tan\theta}{1+\tan^{2}\theta}$
trigonometry
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Best Answer
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.