Solve the equation for $x$: $\sin^{-1} \left( \frac{5}{x} \right) + \sin^{-1} \left( \frac{12}{x} \right) = \frac{\pi}{2} \quad (x \neq 0)$.

$x = 13$

The correct answer is Option (1) → $x = 13$ ##

Given equation can be written as:

$\sin^{-1} \left( \frac{12}{x} \right) = \frac{\pi}{2} - \sin^{-1} \left( \frac{5}{x} \right)$

$\Rightarrow \sin^{-1} \left( \frac{12}{x} \right) = \cos^{-1} \left( \frac{5}{x} \right)$

$∴\sin^{-1} \left( \frac{12}{x} \right) = \sin^{-1} \left( \frac{\sqrt{x^2 - 25}}{x} \right)$

$\Rightarrow \frac{12}{x} = \frac{\sqrt{x^2 - 25}}{x}$

$\Rightarrow x^2 - 25 = 144$

$\Rightarrow x = \pm 13$

Since $x = -13$ does not satisfy the given equation,

$∴$ Required solution is $x = 13$.