Explanation

(b)
Given, $\sin^{- 1}{\left( \frac{3}{x} \right) = \frac{\pi}{2} - \sin^{- 1}\left( \frac{4}{x} \right)}$
$\Rightarrow \sin^{- 1}{\left( \frac{3}{x} \right) = \cos^{- 1}\left( \frac{4}{x} \right)}$
$\Rightarrow \ \sin^{- 1}{\left( \frac{3}{x} \right) = \sin^{- 1}\left( \frac{\sqrt{x^{2} - 16}}{x} \right)}$
$\Rightarrow \ \frac{3}{x} = \frac{\sqrt{x^{2} - 16}}{x}$
$\Rightarrow \ \ x = \pm 5$
$\therefore\ \ \ \ x = 5$
[$\because - 5$ not satisfies the given equation]