$\text{ If }{\sin }^{-1}\left(\mathrm{x}-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^3}{4}-\dots \right)+{\cos }^{-1}\left({\mathrm{x}}^2-\frac{{\mathrm{x}}^4}{2}+\frac{{\mathrm{x}}^6}{4}-\dots \right)=\frac{\mathrm{\pi }}{2}\text{ for }0<\left|\mathrm{x}\right|<\sqrt{2},$ then  100x equals

$$\begin{gathered} \text{ Since }{\sin }^{-1}\mathrm{x}+{\cos }^{-1}\mathrm{x}=\frac{\mathrm{\pi }}{2}\text{ for }\left|\mathrm{x}\right|\le 1\text{. } \\ \therefore \mathrm{x}-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^3}{4}-\dots ={\mathrm{x}}^2-\frac{{\mathrm{x}}^4}{2}+\frac{{\mathrm{x}}^6}{4}-\dots \\ \Rightarrow \frac{\mathrm{x}}{1+\frac{\mathrm{x}}{2}}=\frac{{\mathrm{x}}^2}{1+\frac{{\mathrm{x}}^2}{2}}(0<|\mathrm{x}|<\sqrt{2}) \\ \Rightarrow \frac{\mathrm{x}}{2+\mathrm{x}}=\frac{{\mathrm{x}}^2}{2+{\mathrm{x}}^2}\Rightarrow 2\mathrm{x}+{\mathrm{x}}^3=2{\mathrm{x}}^2+{\mathrm{x}}^3 \end{gathered}$$
⇒ x² = x ⇒ x = 0, 1
But x ≠ 0, ∴ x = 1.