If ^ - 1 (x - x ^ 2 2 + x ^ 3 4 - ) + ^ - 1 ( x ^ 2 - x ^ 4 2 + x ^...

Since, $\sin^{- 1} x + \cos^{- 1} x = \dfrac{\pi}{2}$ for \left|x\right| \le 1 .$<br />$∴ x - \dfrac{{x}^{2}}{2} + \dfrac{{x}^{3}}{4} - \ldots = {x}^{2} - \dfrac{{x}^{4}}{2} + \dfrac{{x}^{6}}{4} - \ldots$<br />$⇒ \dfrac{x}{1 + \dfrac{x}{2}} = \dfrac{{x}^{2}}{1 + \dfrac{{x}^{2}}{2}}\left(0 < \left|x\right| < \sqrt{2}\right)$<br />$⇒ \dfrac{x}{2 + x} = \dfrac{{x}^{2}}{2 + {x}^{2}} ⇒ 2x + {x}^{3} = 2{x}^{2} + {x}^{3}$<br />$⇒ {x}^{2} = x ⇒ x = 0, 1$<br />But${x}^{2} = x \ne 0, ∴ x = 1 .