Explanation

(c)
Given, $\sin^{- 1}{x + \sin^{- 1}{(1 - x) = \cos^{- 1}x}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \sin^{- 1}{(1 - x) = \frac{\pi}{2} - \sin^{- 1}{x - \sin^{- 1}x}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{\pi}{2} - 2\sin^{- 1}x$
$\Rightarrow \sin^{- 1}{(1 - x) = \sin^{- 1}{1 - \sin^{- 1}{2x\sqrt{1 - x^{2}}}}}$
$\Rightarrow \ \sin^{- 1}{(1 - x) = \sin^{- 1}{\lbrack 1\sqrt{1 - 4x^{2}\left( 1 - x^{2} \right)} - 0\rbrack}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ (1 - x) = \sqrt{1 - 4x^{2} + 4x^{4}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 - x = 1 - 2x^{2}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ 2x^{2} - x = 0$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \in \left\{ 0,\frac{1}{2} \right\}$