Question

Solve this?

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

`a. 1`

`b. 1/2`

`c. 0`

`d.-1`

Answer 1

`a.` `1`

Explanation:

`sin^-1 theta+cos^-1theta=pi/2`
You have:
`sin^-1(x-x^2/2+x^3/4-...)+cos^-1(x^2-x^4/2+x^6/4-...)=pi/2`

Thus, we can say,
`(x-x^2/2+x^3/4-...)=(x^2-x^4/2+x^6/4-...)`
[because `sin^-1 theta+cos^-1theta=pi/2`; so `theta` is the common or same angle]

From the equation, we understand:
`x=x^2,` `x^2=x^4`,`x^3=x^6,` and so on.
These can be possible only when `(x=1)` or when `(x=0)`.

`color(blue) (0< x < sqrt2.`
Thus, as `x>0`, the only possible value of `x` is 1.