Tangent of acute angle between the curves $\mathrm{y}=\left|{\mathrm{x}}^2-1\right| \mathrm{and} \mathrm{y}=\sqrt{7-{\mathrm{x}}^2}$ at their points of intersection is

$\begin{array}{l}\mathrm{Solving} \mathrm{y}=\left|{\mathrm{x}}^2-1\right| \mathrm{and} \mathrm{y}=\sqrt{7-{\mathrm{x}}^2}, \mathrm{we} \mathrm{have}\\ \left|{\mathrm{x}}^2-1\right|=\sqrt{7--{\mathrm{x}}^2}\\ \mathrm{or} {\mathrm{x}}^4-2{\mathrm{x}}^2+1=7-{\mathrm{x}}^2\\ \mathrm{or} {\mathrm{x}}^4-{\mathrm{x}}^2-6=0\\ \mathrm{or} \left({\mathrm{x}}^2-3\right)\left({\mathrm{x}}^2+2\right)=0\\ \mathrm{or} \mathrm{x}=\pm \sqrt{3}\end{array}$

Points of intersection of the curves $\mathrm{y}=\left|{\mathrm{x}}^2-1\right|$ and 

$\mathrm{y}=\sqrt{7-{\mathrm{x}}^2} \mathrm{are} \left(\pm \sqrt{3,}2\right).$

Since both the curves are symmetrical about the y-axis, points of intersection are also symmetrical.

$\begin{array}{l}\mathrm{Now}, \mathrm{y}={\mathrm{x}}^2-1 \mathrm{or} \frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{x}\\ \mathrm{or} {\mathrm{m}}_1={\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|}_{\left(\sqrt{3},2\right)}=2\sqrt{3}\\ \mathrm{and} \mathrm{y}=\sqrt{7-{\mathrm{x}}^2} \mathrm{or} \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{x}}{\mathrm{y}}\\ \mathrm{or} {\mathrm{m}}_2={\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|}_{\left(\sqrt{3},2\right)}=-\frac{\sqrt{3}}{2}\\ \mathrm{Therefore}, \mathrm{tan\theta }=\left|\frac{5\sqrt{3}}{4}\right|\end{array}$