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

1

We have,

$sin^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}-\frac{x^4}{8}+.....\right) +cos^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}.....\right) = \frac{\pi}{2}$

$⇒ sin^{-1}\left(\frac{x}{1+\frac{x}{2}}\right)+cos^{-1}\left(\frac{x^2}{1+\frac{x}{2}}\right)=\frac{\pi}{2}$

$⇒ sin^{-1}\left(\frac{2x}{2+x}\right) +cos^{-1}\left(\frac{2x^2}{2+x^2}\right) =\frac{\pi}{2}$

$⇒ sin^{-1}\left(\frac{2x}{2+x}\right) =sin^{-1}\left(\frac{2x^2}{2+x^2}\right)$

$⇒ \frac{2x}{2+x}=\frac{2x^2}{2+x^2}⇒\frac{x}{2+x}=\frac{x^2}{2+x^2}$

$⇒1+\frac{2}{x}=1+\frac{2}{x^2}⇒\frac{1}{x}=\frac{1}{x^2}⇒x = x^2 ⇒ x = 1 $    $[∵ x > 0]$