If ${\sin }^{-1}\left(\mathrm{x}-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^3}{4}-\dots \right)+{\cos }^{-1}\left({\mathrm{x}}^2-\frac{{\mathrm{x}}^4}{2}+\frac{{\mathrm{x}}^6}{4}-\dots \right)$ $=\frac{\mathrm{\pi }}{2},\text{ for }0<\left|\mathrm{x}\right|<\sqrt{2},$ then x values 

We know that, ${\sin }^{-1}\left(\mathrm{\alpha }\right)+{\cos }^{-1}\left(\mathrm{\alpha }\right)=\frac{\mathrm{\pi }}{2}$

Therefore, $\alpha$ should be equal in both functions

$\begin{array}{l}\therefore \mathrm{x}-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^3}{4}-\dots ={\mathrm{x}}^2-\frac{{\mathrm{x}}^4}{2}+\frac{{\mathrm{x}}^6}{4}-\dots \\ \Rightarrow \frac{\mathrm{x}}{1+\frac{\mathrm{x}}{2}}=\frac{{\mathrm{x}}^2}{1+\frac{{\mathrm{x}}^2}{2}}\\ \Rightarrow \frac{x}{{\displaystyle \frac{2+x}{2}}}=\frac{{\mathrm{x}}^2}{\frac{2+{\mathrm{x}}^2}{2}}\end{array}$

$\begin{array}{ccc}\Rightarrow & \frac{2\mathrm{x}}{2+\mathrm{x}}=\frac{2{\mathrm{x}}^2}{2+{\mathrm{x}}^2}& \\ \Rightarrow & 2\mathrm{x}\left(2+{\mathrm{x}}^2\right)=2{\mathrm{x}}^2(2+\mathrm{x})& \\ \Rightarrow & 4\mathrm{x}+2{\mathrm{x}}^3=4{\mathrm{x}}^2+2{\mathrm{x}}^3& \\ \Rightarrow & \mathrm{x}\left(4+2{\mathrm{x}}^2-4\mathrm{x}-2{\mathrm{x}}^2\right)=4{\mathrm{x}}^2+2{\mathrm{x}}^3& \\ \Rightarrow & \text{ Either }\mathrm{x}=0\text{ or }4-4\mathrm{x}=0& \\ \Rightarrow & \mathrm{x}=0\text{ or }\mathrm{x}=1& \\ \because & 0<\left|\mathrm{x}\right|<\sqrt{2}& \\ \therefore & \mathrm{x}=1\text{ and }\mathrm{x}\ne 0& \end{array}$