If `sin^(- 1)(x-(x^2)/2+(x^3)/4-.....)+cos^(- 1)(x^2-(x^4)/2+(x^6)/4-.....)=pi/2` for `0 lt |x| lt sqrt2` then `x=`

Options

A. `1/2`
B. 1
C. `-1/2`
D. `-1`

Answer

B

Explanation

We have
`sin^(-1)(x-(x^(2))/(22)+(x^(3))/(4)-(x^(4))/(8)+..)+cos^(-1)(x^(2))-(x^(4))/(2)+(x^(6))/(2)..)=(pi)/(2)`
`rarr sin^(-1)(x)/(1+(x)/(2))+cos^(-1)((2x^(2))/(2+x^(2))=(pi)/(2)`
`rarr sin^(-1)((2x)/(2+x))=sin^(1)((2x^(2))/(2+x^(2))`
`rarr (2x)/(2+x)=(2x^(2))/(2+x^(2))rarr(x)/(x+2)=(x^(2))/(2+x^(2))`
`rarr sin^(-1)(2x)/(2+x)=(2x^(2))/(2+x^(2))rarr (x)/(x+2)=(x^(2))/(x2+x^(2))`
`rarr 1+(2)/(x)=1+(2)/(x^(2))rarr(1)/(x)=(1)/(x^(2))rarrx=x^(2)rarrx=1`