## Calculus: Early Transcendentals 8th Edition

$tan2\theta=\frac{2tan\theta}{1-tan^{2}\theta}$
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, $tan2\theta=\frac{2tan\theta}{1-tan^{2}\theta}$