If ${\sin }^{-1}(x^2-\frac{x^4}{3}+\frac{x^6}{9}\mathrm{......})+{\cos }^{-1}(x^4-\frac{x^8}{3}+\frac{x^{12}}{9}\mathrm{.....})=\frac{\pi }{2},$ where $0\le \left|x\right|<\sqrt{3},$ then number of values of ‘x’ is equal to

${\sin }^{-1}\underset{x}{\underbrace{(x^2-\frac{x^4}{3}+\frac{x^6}{9}\mathrm{.......})}}+{\cos }^{-1}\underset{y}{\underbrace{(x^4-\frac{x^8}{3}+\frac{x^{12}}{9}\mathrm{......})}}=\frac{\pi }{2}$

$\Rightarrow X=Y\Rightarrow \frac{x^2}{1+\frac{x^2}{3}}=\frac{x^4}{1+\frac{x^4}{3}}\Rightarrow \frac{3}{3+x^2}=\frac{3x^2}{3+x^4}\Rightarrow 9+3x^4=9x^2+3x^4\Rightarrow x^2=1$

Thus, $x=0,1 or -1$

Hence, number of values is equal to 3.