Find intersection points in the first quadrant ($x>0,y>0$): $|x^2-1|=1⟹x^2-1=1$ or $x^2-1=-1$. $x^2=2⟹x=\sqrt{2}$. (Note: $x^2=0$ gives intersection at $x=0$, but 'first quadrant' implies $x>0$, so we exclude $x=0$. The actual intersections are $(0,1)$ and $(\sqrt{2},1)$).

Find slopes at $x=\sqrt{2}$: Curve 1: $y=x^2-1$ (since $x^2>1$ near $\sqrt{2}$). $dy/dx=2x$. At $x=\sqrt{2},m_1=2\sqrt{2}$. Curve 2: $y=1$. Constant function. $dy/dx=0$. So $m_2=0$.

Calculate angle: $\tan \theta =|\frac{m_1-m_2}{1+m_1{m}_2}|=|\frac{2\sqrt{2}-0}{1+0}|=2\sqrt{2}$. Thus, $\theta ={\tan }^{-1}(2\sqrt{2})$.