The value of $\tan \left({\sin }^{-1}\left(\cos \left({\sin }^{-1}\mathrm{x}\right)\right)\right)=\tan \left({\cos }^{-1}\left(\sin \left({\cos }^{-1}\mathrm{x}\right)\right)\right).\text{ where }\mathrm{x}\in (0,1)$ is equal to

$\begin{array}{l}\tan \left({\sin }^{-1}\left(\cos \left({\cos }^{-1}\sqrt{1-{\mathrm{x}}^2}\right)\right)\right)=\tan \left({\sin }^{-1}\sqrt{1-x^2}\right)=\tan \left({\tan }^{-1}\frac{\sqrt{1-x^2}}{x}\right)=\frac{\sqrt{1-x^2}}{x}\\ \tan \left({\cos }^{-1}\left(\sin \left({\sin }^{-1}\sqrt{1-{\mathrm{x}}^2}\right)\right)\right)=\tan \left({\cos }^{-1}\sqrt{1-{\mathrm{x}}^2}\right)=\tan \left({\tan }^{-1}\frac{x}{\sqrt{1-x^2}}\right)=\frac{x}{\sqrt{1-x^2}}\end{array}$
$$\begin{gathered} \frac{\sqrt{1-x^2}}{x}=\frac{x}{\sqrt{1-x^2}} \\ \Rightarrow x^2=1-x^2 \\ \Rightarrow 2x^2=1\Rightarrow x^2=\frac{1}{2}\Rightarrow x=\pm \sqrt{\frac{1}{2}} \end{gathered}$$