1. The solution set of the inequality $\arccos x<\arctan x$ is
$\qquad$
$$
\text { II. 1. }\left\{x \left\lvert\, \sqrt{\frac{\sqrt{5}-1}{2}}0$, both sides of the inequality are acute angles. The original inequality is equivalent to
$\operatorname{tan}(\arccos x)\frac{\sqrt{5}-1}{2}$ or $x^{2}\sqrt{\frac{\sqrt{5}-1}{2}}$.
Also, since $x \in(0,1]$,
$$
\therefore \sqrt{\frac{\sqrt{5}-1}{2}}<x \leqslant 1 \text {. }
$$
$$