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

Solution :
$${\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - .....} \right) + {\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - .....} \right) = \frac{\pi }{2}$$
This is true only when $$x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - ..... = {x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4}.....$$
$$ \Rightarrow \frac{x}{{1 + \frac{x}{2}}} = \frac{{{x^2}}}{{1 + \frac{{{x^2}}}{2}}}$$
(Common ratios are $$ - \frac{x}{2}\,\,\& \,\, - \frac{{{x^2}}}{2}\,\,\& \,\,\left| {{\text{common ratios}}} \right| < 1,\,$$       in the given interval)
$$\eqalign{ & \frac{{2x}}{{2 + x}} = \frac{{2{x^2}}}{{2 + {x^2}}} \cr & \Rightarrow x = 0{\text{ or }}x = 1 \cr & \Rightarrow x = 1, \cr} $$
{ $$x$$ cannot be zero as $$0 < \left| x \right| < \sqrt 2 $$   }.