# Solve this?

## If ${\sin}^{- 1} \left(x - {x}^{2} / x + {x}^{3} / 4. \ldots .\right) + {\cos}^{- 1} \left({x}^{2} - {x}^{4} / 2 + {x}^{6} / 4. \ldots .\right) = \frac{\pi}{2}$ for $0 < x < \sqrt{2}$ then x equals $a . 1$ $b . \frac{1}{2}$ $c . 0$ $d . - 1$

Mar 20, 2018

$a .$ $1$

#### Explanation:

${\sin}^{-} 1 \theta + {\cos}^{-} 1 \theta = \frac{\pi}{2}$
You have:
${\sin}^{-} 1 \left(x - {x}^{2} / 2 + {x}^{3} / 4 - \ldots\right) + {\cos}^{-} 1 \left({x}^{2} - {x}^{4} / 2 + {x}^{6} / 4 - \ldots\right) = \frac{\pi}{2}$

Thus, we can say,
$\left(x - {x}^{2} / 2 + {x}^{3} / 4 - \ldots\right) = \left({x}^{2} - {x}^{4} / 2 + {x}^{6} / 4 - \ldots\right)$
[because ${\sin}^{-} 1 \theta + {\cos}^{-} 1 \theta = \frac{\pi}{2}$; 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 $\left(x = 1\right)$ or when $\left(x = 0\right)$.

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