If ${\cos }^{-1}\left(\frac{3}{5}\right)-{\sin }^{-1}\left(\frac{4}{5}\right)={\cos }^{-1}x$, then $x=$

Let $\theta ={\sin }^{-1}\left(\frac{4}{5}\right)$. This means $\sin \theta =\frac{4}{5}$.In a right-angled triangle, $\cos \theta =\sqrt{1-{\sin }^2\theta }=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$.So, ${\sin }^{-1}\left(\frac{4}{5}\right)={\cos }^{-1}\left(\frac{3}{5}\right)$.Substituting this back into the equation:
$${\cos }^{-1}\left(\frac{3}{5}\right)-{\cos }^{-1}\left(\frac{3}{5}\right)={\cos }^{-1}x$$
$$0={\cos }^{-1}x$$
$$x=\cos (0)=1$$