Let $\mathrm{S}=\left\{\mathrm{x}\in \mathrm{R}:0<\mathrm{x}<1\text{ and }2{\tan }^{-1}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)={\cos }^{-1}\left(\frac{1-{\mathrm{x}}^2}{1+{\mathrm{x}}^2}\right)\right\}$
If $n\left(S\right)$ denotes the number of elements in $S$ then :

$0<\mathrm{x}<1$
If $\mathrm{x}=\mathrm{Tan}\mathrm{\theta },\mathrm{\theta }\in \left(0,\frac{\mathrm{\pi }}{4}\right),2\mathrm{\theta }\in \left(0,\frac{\mathrm{\pi }}{2}\right)$
$\begin{array}{l}2{\mathrm{Tan}}^{-1}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)={\cos }^{-1}\left(\frac{1-{\mathrm{x}}^2}{1+{\mathrm{x}}^2}\right)\\ \Rightarrow 2{\mathrm{Tan}}^{-1}\left(\mathrm{Tan}\left(\frac{\mathrm{\pi }}{4}-\mathrm{\theta }\right)\right)={\cos }^{-1}\left(\frac{1-{\mathrm{Tan}}^2\mathrm{\theta }}{1+{\mathrm{Tan}}^2\mathrm{\theta }}\right)\\ \Rightarrow 2\left(\frac{\mathrm{\pi }}{4}-\mathrm{\theta }\right)={\cos }^{-1}(\cos 2\mathrm{\theta })\\ \Rightarrow \frac{\mathrm{\pi }}{2}-2\mathrm{\theta }=2\mathrm{\theta }\\ \Rightarrow \mathrm{\theta }=\frac{\mathrm{\pi }}{8}<\frac{1}{2}\therefore \mathrm{n}\left(\mathrm{s}\right)=1\end{array}$