Convert $\sin\left(2\cos^{-1}\left(\cot\left(2 \tan^{-1}x\right)\right)\right)$ into an algebraic function 
Convert trigonometric function into algebraic function.
  $$\sin\left(2\cos^{-1}\left(\cot\left(2 \tan^{-1}x\right)\right)\right)$$

My approach is as follows:
$sin2\theta$ is to be calculated
$$\tan^{-1}x=\gamma \tag{1}$$
Hence 
$$\begin{align}
\cos\theta&=\cot2\gamma \tag{2}\\[4pt]
\cos\theta&=\frac{\cot^2\gamma-1}{2\cot\gamma} \tag{3}\\[4pt]
\cos\theta&=\frac{1-x^2}{2x} \tag{4}
\end{align}$$
The value of $\sin\theta$ is coming in negative.
 A: If $\cos^{-1}y=u,\cos u=y,0\le y\le\pi$
If $\sin(2\cos^{-1}y)=\sin2u=2\sin u\cos u=\begin{cases}2u\sqrt{1-u^2} &\mbox{if }0\le y\le\dfrac\pi2 \\ 
-2u\sqrt{1-u^2} & \mbox{if } y>\dfrac\pi2 \end{cases} $
Here $u=\dfrac{1-x^2}{2x}$
For real $\cos^{-1}y,-1\le\cot(2\tan^{-1}x)\le1$
$\iff\dfrac\pi4\le2\tan^{-1}x\le\dfrac{3\pi}4$
$\iff\tan\dfrac\pi8\le x\le\tan\dfrac{3\pi}8$
A: I firstly apologise about how this is written, I really ought to learn how to type this stuff properly. Secondly, if there's any mistakes please do say! 
$$\tan(2x) = 2\tan(x)/1-\tan^2(x)$$
so $$\cot(2x) = (1-\tan^2(x))/2\tan(x)$$
swapping $x = \arctan(x)$
$$\cot(2\arctan(x)) = (1-x^2)/2x$$
Okay, that bit is done,
Let $(1-x^2)/2x = Y$ for simplicity...
So we want to evaluate 
$$\sin(2\arccos(Y))$$
Well, $$\sin(2x)= 2\sin(x)\cos(x)$$
this gives us
$$2\sin(\arccos(Y))(Y)$$
So we just want to now evaluate $$2\sin(\arccos(Y))$$ 
Well $$\sin^2(x)+\cos^2(x)=1$$ 
so,
$$\sin(x) = \sqrt {(1 - \cos^2(x)})$$ 
$$\sin(\arccos(y)) = \sqrt {(1-Y^2)}$$
$$2\sin(\arccos(y)) = \sqrt{2(1-Y^2)}$$
So the answer is $$2Y \sqrt {(1-Y^2)}$$
where $Y=(1-x^2)/2x$.
A: \begin{align}
   \cot\left(2 \tan^{-1}x\right)
   &= \dfrac{1}{\tan\left(2 \tan^{-1}x\right)} \\
   &= \dfrac{1-\tan^2(\tan^{-1}x)}{2\tan(\tan^{-1}x)} \\
   &= \dfrac{1-x^2}{2x} \\
\hline
   \sin\left(2\cos^{-1}\left(\cot\left(2 \tan^{-1}x\right)\right)\right) 
   &= 2\sin\left(\dfrac{ 1-x^2}{2x} \right) \cos\left( \dfrac{1-x^2}{2x} \right)
\end{align}
