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

Let $A=\sin^{-1}\left(\frac{5}{x}\right)$ and $B=\sin^{-1}\left(\frac{12}{x}\right)$. Given $A+B=\frac{\pi}{2}$, so $B=\frac{\pi}{2}-A$. Hence $\sin B=\cos A$. So $$\frac{12}{x}=\sqrt{1-\frac{25}{x^2}}.$$ Squaring, $$\frac{144}{x^2}=1-\frac{25}{x^2}.$$ Thus $$\frac{169}{x^2}=1.$$ Therefore $$x^2=169.$$ Since $\frac{5}{x}$ and $\frac{12}{x}$ must both lie in the domain of inverse sine and their inverse sines add to $\frac{\pi}{2}$, we take the positive value, so $$x=13.$$ On checking the options, this matches option $\text{C}$.