Prove the following identities.
$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$

Verified step by step guidance

1. Start by recalling the double angle identity for tangent: $\tan (2\theta )=\frac{2\tan (\theta )}{1-{\tan }^2(\theta )}$. This is a standard trigonometric identity.
2. To prove this identity, let's use the sine and cosine double angle identities: $\sin (2\theta )=2\sin (\theta )\cos (\theta )$ and $\cos (2\theta )={\cos }^2(\theta )-{\sin }^2(\theta )$.
3. Express $\tan (2\theta )$ in terms of sine and cosine: $\tan (2\theta )=\frac{\sin (2\theta )}{\cos (2\theta )}$. Substitute the double angle formulas: $\tan (2\theta )=\frac{2\sin (\theta )\cos (\theta )}{{\cos }^2(\theta )-{\sin }^2(\theta )}$.
4. Now, express $\tan (\theta )$ as $\frac{\sin (\theta )}{\cos (\theta )}$. Substitute this into the expression: $\tan (2\theta )=\frac{2\left(\frac{\sin (\theta )}{\cos (\theta )}\right)}{1-{\left(\frac{\sin (\theta )}{\cos (\theta )}\right)}^2}$.
5. Simplify the expression: $\tan (2\theta )=\frac{2\tan (\theta )}{1-{\tan }^2(\theta )}$. This matches the given identity, thus proving it.