If sin ^{-1}left(x-frac{x^2}{2}+frac{x^3}{4}- | vedclass

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(B) The given equation is $\sin ^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}-\ldots \infty\right) + \cos ^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\ldots

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(B) The given equation is $\sin ^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}-\ldots \inftyight) + \cos ^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\ldots \inftyight)=\frac{\pi}{2}$.Both series are infinite geometric progressions with common ratios $-x/2$ and $-x^2/2$ respectively.The sum of an infinite $G$.$P$. is $S = \frac{a}{1-r}$.For the first series,$a=x$ and $r=-x/2$,so the sum is $\frac{x}{1-(-x/2)} = \frac{x}{1+x/2} = \frac{2x}{2+x}$.Wait,the series given is $x - x^2/2 + x^3/4 - \dots$,which is a $G$.$P$. with $a=x$ and $r=-x/2$. The sum is $\frac{x}{1-(-x/2)} = \frac{2x}{2+x}$.Similarly,for the second series $x^2 - x^4/2 + x^6/4 - \dots$,$a=x^2$ and $r=-x^2/2$. The sum is $\frac{x^2}{1-(-x^2/2)} = \frac{2x^2}{2+x^2}$.Using the property $\sin ^{-1}(u) + \cos ^{-1}(v) = \pi/2$,we must have $u=v$.Thus,$\frac{2x}{2+x} = \frac{2x^2}{2+x^2}$.Since $x 
eq 0$,we divide by $2x$: $\frac{1}{2+x} = \frac{x}{2+x^2}$.$2+x^2 = x(2+x) \implies 2+x^2 = 2x+x^2$.$2x = 2 \implies x=1$.

If $\sin ^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}-\ldots \infty\right) + \cos ^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\ldots \infty\right)=\frac{\pi}{2}$ and $0 < x < \sqrt{2}$,then $x$ is equal to

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