If ${\sin }^{-1}\left(\frac{x}{5}\right)+{\text{cosec}}^{-1}\left(\frac{5}{4}\right)=\frac{\pi }{2}$, then a value of $x$ is:

Given: ${\sin }^{-1}\left(\frac{x}{5}\right)+{\text{cosec}}^{-1}\left(\frac{5}{4}\right)=\frac{\pi }{2}$ We know that ${\text{cosec}}^{-1}\left(\frac{5}{4}\right)={\sin }^{-1}\left(\frac{4}{5}\right)$.So, ${\sin }^{-1}\left(\frac{x}{5}\right)+{\sin }^{-1}\left(\frac{4}{5}\right)=\frac{\pi }{2}$ $
${\sin }^{-1}\left(\frac{x}{5}\right)=\frac{\pi }{2}-{\sin }^{-1}\left(\frac{4}{5}\right)$$Since$ \sin^{-1}\theta + \cos^{-1}\theta = \frac{\pi}{2}$,wehave:$${\sin }^{-1}\left(\frac{x}{5}\right)={\cos }^{-1}\left(\frac{4}{5}\right)$$Let$ \cos^{-1}\left(\frac{4}{5}\right) = \theta \implies \cos\theta = \frac{4}{5}$.Then$ \sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.Thus,$ \theta = \sin^{-1}\left(\frac{3}{5}\right)$.Substitutingback:$${\sin }^{-1}\left(\frac{x}{5}\right)={\sin }^{-1}\left(\frac{3}{5}\right)$$Comparingbothsides,weget$ x = 3$.