Answer
$tan2\theta=\frac{2tan\theta}{1-tan^{2}\theta}$
Work Step by Step
We need to prove the identity $tan2\theta=\frac{2tan\theta}{1-tan^{2}\theta}$
$tan2\theta =tan(\theta+\theta)$
$tan(\theta+\theta)=\frac{tan\theta+tan\theta}{1-tan\theta\times tan\theta}$ [sum identity for tangent]
Thus,
$tan(\theta+\theta)=\frac{2tan\theta}{1-tan^{2}\theta}$
Hence, $tan2\theta=\frac{2tan\theta}{1-tan^{2}\theta}$
