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# If $\sin^{-1}\big(x-\large\frac{x^2}{2}+\frac{x^3}{4}-........)$$+\cos^{-1}(x^2-\large\frac{x^4}{2}+\frac{x^6}{4}.......)=\large\frac{\pi}{2}.For 0<\mid x\mid<\sqrt 2 then x equals \begin{array}{1 1}(a)\;\large\frac{1}{2}&(b)\;1\\(c)\;-\large\frac{1}{2}&(d)\;-1\end{array} Can you answer this question? ## 1 Answer 0 votes Toolbox: • Sum of infinite G.P.=\large\frac{a}{1-r} where a=1^{st}\:term and r=common\: ratio. Given: \sin^{-1}\big(x-\large\frac{x^2}{2}+\frac{x^3}{4}-........)$$+\cos^{-1}(x^2-\large\frac{x^4}{2}+\frac{x^6}{4}.......)=\large\frac{\pi}{2}$
$x-\large\frac{x^2}{2}+\frac{x^3}{4}-.......$ is an infinite G.P.
Sum of this series $=\large\frac{x}{1+x/2}$$=\large\frac{2x}{2+x} Similarly x^2-\large\frac{x^4}{2}+\frac{x^6}{4}....... is also an infinite G.P. Sum of this series =\large\frac{x^2}{1+x^2/2}$$=\large\frac{2x^2}{2+x^2}$
Substituting these values in the given equation, we get
$\sin^{-1}\big(\large\frac{x}{1+x/2}\big)$$+\cos^{-1}\big(\large\frac{x^2}{1+x^2/2}\big)=\frac{\pi}{2} \Rightarrow\:\sin^{-1}\big(\large\frac{2x}{2+x}\big)=\frac{\pi}{2}$$-\cos^{-1}\big(\large\frac{2x^2}{2+x^2}\big)$
$\qquad\quad\quad\quad\;\;=\sin^{-1}\big(\large\frac{2x^2}{2+x^2}\big)$
$\Rightarrow\:\large\frac{2x}{2+x}=\frac{2x^2}{2+x^2}$