Simplifying $\sin(2\tan^{-1} x)$

Hint: $\cos^2(\alpha) = \dfrac{\cos^2(\alpha)}{\cos^2(\alpha) + \sin^2(\alpha)} $

Making a right-triangle picture can be helpful too, either for getting to the result quickly or checking up on the analytical result. With $\tan \alpha = x = \frac{x}{1}$, we would have $x$ for the leg opposite $\alpha$ and $1$ for the adjacent leg; the hypotenuse is then $\sqrt{x^2 + 1}$ . We wish to evaluate $\sin 2\alpha$ , which is thus

$$2 \sin \alpha \cos \alpha = 2 \cdot \frac{x}{\sqrt{x^2 + 1}} \cdot \frac{1}{\sqrt{x^2 + 1}} = \frac{2x}{x^2 + 1} .$$

Hint: What's the reciprocal of $\cos^2 \alpha$?