If $\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\left(\frac{5}{x}\right)+\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\left(\frac{12}{x}\right)=\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\left(\frac{2}{x}\right)+\mathrm{c}\mathrm{o}{\mathrm{s}}^{-1}\left(\frac{2}{x}\right)$, then the value of $x$ is equal to

$\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\left(\frac{5}{x}\right)$ and $\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\left(\frac{12}{x}\right)$ are both defined when $x>12.$ Therefore
$\mathrm{c}\mathrm{o}{\mathrm{s}}^{-1}\left(\sqrt{1-\frac{25}{x^2}}\right)+\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\left(\frac{12}{x}\right)=\frac{\pi }{2}$ $\left\{\because \mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\frac{2}{x}+\mathrm{c}\mathrm{o}{\mathrm{s}}^{-1}\frac{2}{x}=\frac{\pi }{2}\right\}$
$\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\left(\frac{12}{x}\right)=\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}\sqrt{1-\frac{25}{x^2}}$
On comparing, we get,
$\frac{12}{x}=\sqrt{1-\frac{25}{x^2}}$

Therefore, $144=x^2-25$ or $x=13$