If $∆={\mathrm{a}}^2-{(\mathrm{b}-\mathrm{c})}^2,$ is the area of the triangle ABC, then $\tan \mathrm{A}=$

$$\begin{gathered} Given ∆=a^2-{(b-c)}^2 \Rightarrow ∆=(a+b-c)(a+c-b) \\ \Rightarrow ∆=2(s-c)2(s-b) \\ \Rightarrow \frac{(s-b)(s-c)}{∆}=\frac{1}{4} \\ \Rightarrow \tan \left(\frac{A}{2}\right)=\frac{1}{4} \\ Now \tan A=\frac{2\tan {\displaystyle \frac{A}{2}}}{1-{\tan }^2\frac{A}{2}} \\ =\frac{2\left({\displaystyle \frac{1}{4}}\right)}{1-{\left({\displaystyle \frac{1}{4}}\right)}^2} \\ =\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{15}{16}}} \\ =\frac{8}{15} \end{gathered}$$