Trigonometry! 64

Let the area of $\Delta ABC$ be denoted by $\Delta$ and its semiperimeter by $s$ .

$\Delta = a^2-(b-c)^2 \\ \Rightarrow \Delta = (a+b-c)(a-b+c) \\ \Rightarrow \Delta = (a+b+c-2c)(a+b+c-2b) \\ \Rightarrow \Delta = (2s-2c)(2s-2b) \\ \Rightarrow \Delta = 4(s-c)(s-b) \\ \Rightarrow \Delta^2 = 16(s-c)^2(s-b)^2 \\ \Rightarrow s(s-a)(s-b)(s-c)= 16(s-c)^2(s-b)^2 \\ \Rightarrow s(s-a)=16(s-b)(s-c) \\ \Rightarrow \dfrac{(s-b)(s-c)}{s(s-a)} = \dfrac{1}{16} \\ \Rightarrow \tan^2 \dfrac{A}{2} = \dfrac{1}{16} \\ \Rightarrow \tan\dfrac{A}{2} = \dfrac{1}{4} \\ \because \tan A = \dfrac{2\tan\dfrac{A}{2}}{1-\tan^2\dfrac{A}{2}} \\ \Rightarrow \tan A = \dfrac{2\times \dfrac{1}{4}}{1-\left(\dfrac{1}{4}\right)^2} = \dfrac{\dfrac{1}{2}}{1-\dfrac{1}{16}} = \dfrac{\dfrac{1}{2}}{\dfrac{15}{16}}= \dfrac{1}{2} \times \dfrac{16}{15}=\dfrac{8}{15} \\ \therefore \tan A =\boxed{\dfrac{8}{15}}$

Thus finally , $\Large a+b=8+15 = \boxed{23}$

Let the area of $\Delta ABC$ be denoted by $\Delta$ .

$\therefore \Delta = a^{2}-(b-c)^{2}$

$\Delta = a^{2}-b^{2}-c^{2}+2bc$

$\Delta=2bc-2bc\left(\frac{b^{2}+c^{2}-a^{2}}{2bc}\right)$

$\frac{1}{2}bc\sin A = 2bc - 2bc\cos A$

$\frac{\sin A}{2}=2(1-\cos A)$

$\sin A=8\sin^{2}\frac{A}{2}$

$2\sin\frac{A}{2}\cos{A}{2}=8\sin^{2}\frac{A}{2}$

$\tan\frac{A}{2}=\frac{1}{4}$

$\therefore \tan A = \frac{2\tan\frac{A}{2}}{1-\tan^{2}\frac{A}{2}}=\frac{\frac{1}{2}}{1-\frac{1}{16}}=\frac{1}{2}\times\frac{16}{15}=\boxed{\frac{8}{15}}$