Question

If $\sin {}^{-1}(x-\frac{x^{}}{2}+\frac{x^3}{4}-...)+{\cos }^{-1}(x^2-\frac{x^4}{2}+\frac{x^6}{4}-...)=\frac{\pi }{2}$ for $0<\left|x\right|<\sqrt{2}$, then $x$ equals

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

Answer 1

Answer: (B)

$1$

${\sin }^{-1}(x-\frac{x}{2}+\frac{x^2}{4}-..\infty )+co{s}^{-1}(x^2-\frac{x^4}{2}+\frac{x^3}{6}+...\infty )$

$={\sin }^{-1}(x-[\frac{\frac{x}{2}}{1+\frac{x^2}{2}}\left]\right)+co{s}^{-1}\left(\frac{x^2}{1+\frac{x^2}{2}}\right)$
$={\sin }^{-1}(x-\frac{x}{x^2+2})+co{s}^{-1}\left(\frac{2x^2}{x^2+2}\right)$
$=\frac{\pi }{2}$
Therefore
$x-\frac{x}{x^2+2}=\frac{2x^2}{x^2+2}$
$x^3+2x-x=2x^2$
$x^3-2x^2+x=0$
$x(x^2-2x+1)=0$
$x\left(\right(x-1{)}^2)=0$
$x=0$ and $x=1$
But $|x|>0$
Hence
$x=1$