${\sin }^{-1}\left(\frac{12}{13}\right)+{\sec }^{-1}\left(\frac{13}{x}\right)=\frac{\pi }{2}$then $x$is

$$\begin{gathered} {\sin }^{-1}\left(\frac{12}{13}\right)+se{c}^{-1}\left(\frac{13}{x}\right)=\frac{\mathrm{\pi }}{2} \\ \Rightarrow {\sin }^{-1}\left(\frac{12}{13}\right)+{\cos }^{-1}\left(\frac{x}{13}\right)=\frac{\mathrm{\pi }}{2}\left(\sin ce {\sin }^{-1}x+{\cos }^{-1}x=\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow \frac{12}{13}=\frac{x}{13} \\ \Rightarrow x=12 \end{gathered}$$