\(a^2 - (b-c)^2 = (a^2 - b^2 - c^2) + 2bc = -2bc\cos A + 2bc = 2bc(1 - \cos A)\)

Also \(\Delta = \frac{1}{2}bc\sin A\). So \(\frac{1}{2}bc\sin A = 2bc(1 - \cos A)\), giving \(\sin A = 4(1 - \cos A)\).

\(2\sin\frac{A}{2}\cos\frac{A}{2} = 8\sin^2\frac{A}{2} \Rightarrow \tan\frac{A}{2} = \frac{1}{4}\)

\(\tan A = \frac{2 \cdot \frac{1}{4}}{1 - \frac{1}{16}} = \frac{1/2}{15/16} = \frac{8}{15}\)

Answer: (B) \(\frac{8}{15}\)