SOLUTION : If $x \leq-1$, then $\sec ^{-1} x>\frac{\pi}{2}$ and $\tan ^1 x \leq-\frac{\pi}{4}<0$
$\therefore \sec ^{-1} x>\tan ^{-1} x$ for all $x \leq-1$
If $x \geq 1$, suppose $\tan ^{-1} x=\theta$, then $\frac{\pi}{4} \leq \theta<\frac{\pi}{2}$ and $x=\tan \theta \quad \therefore \sec \theta=\sqrt{1+\tan ^2 \theta}=\sqrt{1+x^2}$
$\therefore \quad \sec ^{-1} \sqrt{1+x^2}=\sec ^{-1}(\sec \theta)=\theta=\tan ^{-1} x$
thus the inequality becomes $\sec ^{-1} x>\sec ^{-1} \sqrt{1+x^2}$
$\therefore \quad x>\sqrt{1+x^2} \quad$ i.e. $\quad x^2>1+x^2$ which is not possible $\therefore\{x: x \in(-\infty,-1)\}$ is the solution set
