<s>[INST] 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.
 [/INST] 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 $Y=(1-x^2)/2x$.
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