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 $\frac{1}{x^2 + 1}$

$\sin(2arctan(x))$
let $y=arctan(x)$ 
so $x=tan(y)$ 
$sin(2y)=2sin(y)cos(y)=2sin(arctan(x))\cdot cos(arctan(x))$
$sin(arctan(x)) = \frac{x}{\sqrt{1+x^2}}$   $cos(arctan(x))= \frac{1}{\sqrt{1+x^2}}$
$2\cdot\frac{x}{\sqrt{1+x^2}} \cdot \frac{1}{\sqrt{1+x^2}}=\frac{2x}{{1+x^2}}$
