If {sin ^{ - 1}}left( {x - frac{{{x^2}}}{2} + | vedclass

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(B) We know that ${\sin ^{ - 1}}y + {\cos ^{ - 1}}y = \frac{\pi }{2}$ for $|y| \le 1$.Given the equation ${\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2}

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(B) We know that ${\sin ^{ - 1}}y + {\cos ^{ - 1}}y = \frac{\pi }{2}$ for $|y| \le 1$.Given the equation ${\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - \dots} ight) + {\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - \dots} ight) = \frac{\pi }{2}$,it implies that the arguments of the inverse trigonometric functions must be equal.Let $y = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - \dots$. This is a geometric series with first term $a = x$ and common ratio $r = -x/2$. The sum is $\frac{x}{1 - (-x/2)} = \frac{x}{1 + x/2} = \frac{2x}{2 + x}$.Similarly,the second argument is $x^2 - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - \dots$,which is a geometric series with $a = x^2$ and $r = -x^2/2$. The sum is $\frac{x^2}{1 + x^2/2} = \frac{2x^2}{2 + x^2}$.Equating the two: $\frac{2x}{2 + x} = \frac{2x^2}{2 + x^2}$.Since $x 
eq 0$,we can divide by $2x$: $\frac{1}{2 + x} = \frac{x}{2 + x^2}$.Cross-multiplying gives $2 + x^2 = 2x + x^2$,which simplifies to $2 = 2x$,so $x = 1$.

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

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