Question

If ${\sin }^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}-\frac{x^4}{8}+...\right)-\frac{\pi }{6}$ where $\left|x\right|<2$ then the value of $x$ is

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. $-\frac{2}{3}$
• D. $-\frac{3}{2}$

Answer 1

Answer: (A)

We have, ${\sin }^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}-\frac{x^4}{8}+\dots \right)=\frac{\pi }{6}$
$\Rightarrow {\sin }^{-1}\left(\frac{x}{1-\left(\frac{-x}{2}\right)}\right)=\frac{\pi }{6}$ $\left[\because S_{\infty }=\frac{a}{1-r}\right]$
$\Rightarrow {\sin }^{-1}\left(\frac{2x}{2+x}\right)=\frac{\pi }{6}$
$\Rightarrow \frac{2x}{2+x}=\sin \frac{\pi }{6}$
$\Rightarrow \frac{2x}{2+x}=\frac{1}{2}$
$\Rightarrow 4x=2+x\Rightarrow 3x=2$
$\Rightarrow x=\frac{2}{3}$