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.

Use the following facts: 


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

*$\displaystyle\frac{2 \tan\theta}{1+\tan^{2}\theta} = 2 \cdot \frac{\sin\theta}{\cos\theta} \cdot \cos^{2}\theta = 2 \cdot \sin\theta \cdot \cos\theta$
