We know, $2{\tan }^{-1}\left(x\right)={\sin }^{-1}\left(\frac{2x}{1+x^2}\right) ={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)={\sec }^{-1}\left(\frac{1+x^2}{1-x^2}\right)$

$\therefore x=\sin \left(2{\tan }^{-1}2\right) = \sin \left({\sin }^{-1}\left(\frac{2^2}{1+2^2}\right)\right) = \frac{4}{5}$

$y=\cos \left(2{\tan }^{-1}3\right) = \cos \left({\cos }^{-1}\left(\frac{1-3^2}{1+3^2}\right)\right) = \frac{-4}{5}$

$z=\sec \left(2{\tan }^{-1}4\right) \sec \left({\sec }^{-1}\left(\frac{1+4^2}{1-4^2}\right)\right) = \frac{-5}{4}$

$\therefore z<y<x$