Simplifying $\sin(2\tan^{-1} x)$ I've been working on this for a while. The answer in the book is $\frac{2x}{x^2 + 1}$ Here's my workings:
$\sin(2\tan^{-1} x)$
Let $\alpha = \tan^{-1}x \Rightarrow \tan \alpha = x$
$\sin(2\alpha) = 2\sin\alpha\cos\alpha = 2\tan\alpha\cos^2\alpha = 2x\cos^2\alpha$
I'm not sure how to proceed to turn that $cos^2\alpha$ into $\frac{1}{x^2 + 1}$

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