If ${\sin }^{-1}\frac{x}{5}+{\mathrm{cosec}}^{-1}\left(\frac{5}{4}\right)=\frac{\pi }{2},$ then $x=$

We have,
$\begin{array}{l} {\sin }^{-1}\frac{x}{5}+{\mathrm{cosec}}^{-1}\frac{5}{4}=\frac{\pi }{2}\\ \Rightarrow {\sin }^{-1}\frac{x}{5}=\frac{\pi }{2}-{\mathrm{cosec}}^{-1}\frac{5}{4}\\ \Rightarrow {\sin }^{-1}\frac{x}{5}=\frac{\pi }{2}-{\sin }^{-1}\frac{4}{5}\end{array}$

 $\begin{array}{l}\Rightarrow {\sin }^{-1}\frac{x}{5}={\cos }^{-1}\frac{4}{5}\\ \Rightarrow {\sin }^{-1}\left(\frac{x}{5}\right)={\sin }^{-1}\frac{3}{5}\Rightarrow x=3\end{array}$