Find the acute angles between the curves $y=\left|x^2-1\right|$ and $y=\left|x^2-3\right|$ at their point of intersection.

Given $y=\left|x^2-1\right|$ and $y=\left|x^2-3\right|$

There will be two intersection points when $1<x<\sqrt{3}$ and $-\sqrt{3}<x<-1$ .

when $1<x<\sqrt{3}:- y=x^2-1$ and $y=3-x^2$

For both curves to intersect 

$$\begin{gathered} \Rightarrow x^2-1=3-x^2 \\ 2x^2=4 \\ x=\pm \sqrt{2} \end{gathered}$$

$x=-\sqrt{2}$ can't be the point as $1<x<\sqrt{3}$ .

So $x=\sqrt{2}$

Now , $y=x^2-1\Rightarrow y=(\sqrt{2}{)}^2-1=2-1=1$

Slpoe of $y=x^2-1$ at the point $(\sqrt{2}, 1)$ be $m_1$ .

$m_2=\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=-2x\Rightarrow m_2=-2\left(\sqrt{2}\right)=-2\sqrt{2}$

Therefore $m_1=2\sqrt{2}$ and $m_2=-2\sqrt{2}$

We know that, the angle between two lines having slope $m_1$ and $m_2$ (be $\theta$) then;

$\tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$

Here, $m_1=2\sqrt{2}$ and $m_2=-2\sqrt{2}$

$$\begin{gathered} \therefore tan\theta =\left|\frac{-2\sqrt{2}-2\sqrt{2}}{1-\left(2\sqrt{2}\right)\left(2\sqrt{2}\right)}\right| \\ \tan \theta =\left|\frac{-4\sqrt{2}}{-7}\right|\Rightarrow \tan \theta =\frac{4\sqrt{2}}{7} \\ \Rightarrow \theta ={\tan }^{-1}\left(\frac{4\sqrt{2}}{7}\right) \end{gathered}$$

Therefore, The acute angle between the curves $y=\left|x^2-1\right|$ and $y=\left|x^2-3\right|$ at their point of intersection is ${\tan }^{-1}\left(\frac{4\sqrt{2}}{7}\right)$