Question

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

• A. $\frac{1}{2}$
• B. 1
• C. $-\frac{1}{2}$
• D. $-1$

Answer 1

Answer: (B)

Given that, ${\sin }^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}-\dots \infty \right)$
$+{\cos }^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\dots \infty \right)=\frac{\pi }{2}$
$\Rightarrow {\sin }^{-1}\left[\frac{x}{1-\frac{x}{2}}\right]+{\cos }^{-1}\left[\frac{x^2}{1-\frac{x^2}{2}}\right]=\frac{\pi }{2}$
$\Rightarrow {\sin }^{-1}\left[\frac{2x}{2-x}\right]+{\cos }^{-1}\left[\frac{2x^2}{2-x^2}\right]=\frac{\pi }{2}$
$\left[\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}\right]$
$\Rightarrow \frac{2x}{2-x}=\frac{2x^2}{2-x^2}$
$\Rightarrow \left(2-x^2\right)=x(2-x)$
$\Rightarrow 2-x^2=2x-x^2$
$\Rightarrow 2x=2$
$\Rightarrow x=1$