Question

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

• A. $\frac{1}{2}$
• B. $1$
• C. $-\frac{1}{2}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is B $1$

${\sin }^{-1}(x-\frac{x^2}{2}+\frac{x^3}{4}-....\infty )+{\cos }^{-1}(x^2-\frac{x^4}{2}+\frac{x^6}{4}-....\infty )=\frac{\pi }{2}$

$\Rightarrow {\sin }^{-1}\left(\frac{x}{1+\frac{x}{2}}\right)+{\cos }^{-1}\left(\frac{x^2}{1+\frac{x^2}{2}}\right)=\frac{\pi }{2}$

$\Rightarrow {\sin }^{-1}\left(\frac{x}{1+\frac{x}{2}}\right)=\frac{\pi }{2}-{\cos }^{-1}\left(\frac{x^2}{1+\frac{x^2}{2}}\right)={\sin }^{-1}\left(\frac{x^2}{1+\frac{x^2}{2}}\right)$

$\Rightarrow \frac{x}{1+\frac{x}{2}}=\frac{x^2}{1+\frac{x^2}{2}}$

$\Rightarrow x=x^2\Rightarrow x(x-1)=0\Rightarrow x=0,1$ but $0<x<\sqrt{2}$