Question

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

• A. $1/2$
• B. $1$
• C. $-1/2$
• D. $-1$

Answer 1

Answer: (B)

$sin^{-1}\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{4}-...\right)+cos^{-1}\left(x^{2}-\frac{x^{4}}{2}+\frac{x^{6}}{4}-...\right)=\frac{\pi}{2}$
This is true only when $x-\frac{x^{2}}{2}+\frac{x^{3}}{4}-...=x^{2}-\frac{x^{4}}{2}+\frac{x^{6}}{4}.....$
$\Rightarrow \frac{x}{1+\frac{x}{2}}=\frac{x^{2}}{1+\frac{x^{2}}{2}}$
(Common ratios are $-\frac{x}{2} \& -\frac{x^{2}}{2}\& $ |common ratios $| < 1$, in the given interval)
$\frac{2x}{2+x}=\frac{2x^{2}}{2+x^{2}} \Rightarrow x=0$ or $x=1 \Rightarrow x=1,$
$\left\{x\,cannot \,be \,zero \,as\, 0 < |x| <\sqrt{2}\right\}.$