In $\Delta ABC$ , if the area of a triangle satisfies the relation $\Delta =a^2-{(b-c)}^2$, then $\tan \left(\frac{A}{2}\right)=$

$\tan \frac{A}{2}=\frac{\Delta }{s(s-a)}=\frac{a^2-{(b-c)}^2}{\left(\frac{a+b+c}{2}\right)\left(\frac{b+c-a}{2}\right)}=4\left[\frac{{\displaystyle a^2-b^2-c^2+2bc}}{{\displaystyle b^2+c^2+2bc-a^2}}\right]$

$=\frac{4[-\left(\frac{b^2+c^2-a^2}{2bc}\right)+1]}{\frac{b^2+c^2-a^2}{2bc}+1}$

$=\frac{4(1-\cos A)}{1+\cos A}$

$\begin{array}{l}\therefore \tan \frac{A}{2}=4{\tan }^2\frac{A}{2}\\ \tan \frac{A}{2}=\frac{1}{4}\end{array}$