If denotes the acute angle between the curves,

Solution:

Since, the equation of curves are

$y=10-x^{2}$........(1)

$y=2+x^{2}$.........(2)

Adding eqn (1) and (2), we get

$2 y=12 \Rightarrow y=6$

Then, from eqn (1)

$x=\pm 2$

Differentiate equation (1) with respect to $x$

$\frac{d y}{d x}=-2 x \Rightarrow\left(\frac{d y}{d x}\right)_{(2,6)}=-4$ and $\left(\frac{d y}{d x}\right)_{(-2,6)}=4$

Differentiate equation (2) with respect to $x$

$\frac{d y}{d x}=2 x \Rightarrow\left(\frac{d y}{d x}\right)_{(2,6)}=4$ and $\left(\frac{d y}{d x}\right)_{(-2,6)}=-4$

At $(2,6) \tan \theta=\left(\frac{(-4)-(4)}{1+(-4) \times(4)}\right)=\frac{-8}{15}$

At $(-2,6), \tan \theta=\frac{(4)-(-4)}{1+(4)(-4)}=\frac{8}{-15}$

$\therefore|\tan \theta|=\frac{8}{15}$