If $\sin \left[2{\cos }^{-1} \left\{\cot \left(2{\tan }^{-1} x\right)\right\}\right]=0,x>0$, then

The given equation can be written as

$\begin{array}{l}2{\cos }^{-1} \left[\cot \left(2{\tan }^{-1} x\right)\right]=n\pi \\ \Rightarrow {\cos }^{-1} \left[\cot \left(2{\tan }^{-1} x\right)\right]=n\pi /2(n\in \mathbf{I})\end{array}$

Note that n can take only three values 0, 1, 2. 

$\Rightarrow \cot \left(2{\tan }^{-1} x\right)=\cos \frac{n\pi }{2}=\left\{\begin{array}{cc}0& \text{ if }n=1\\ \pm 1& \text{ if }n=0,2\end{array}\right.$

Now, from $\cot \left(2{\tan }^{-1} x\right)=0$, we get

where $k=– 1$ or 0.

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\tan }^{-1} x=\pi /4\end{array}$ or $-\pi /4\Rightarrow x=\pm 1$

and $\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\cot \left(2{\tan }^{-1} x\right)=\pm 1\Rightarrow \tan \left(2{\tan }^{-1} x\right)=\pm 1$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{2x}{1-x^2}=\pm 1\Rightarrow 1-x^2=\pm 2x\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{x}^2\pm 2x-1=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(x\pm 1{)}^2=2\Rightarrow x=\pm \sqrt{2}\pm 1\end{array}$