In triangle ABC, Find $\tan(A)$. 
In triangle ABC, if $(b+c)^2=a^2+16\triangle$, then find $\tan(A)$ . Where $\triangle$ is the area and a, b , c are the sides of the triangle.


$\implies b^2+c^2-a^2=16\triangle-2bc$
In triangle ABC, $\sin(A)=\frac{2\triangle}{bc}$, and $\cos(A)=\frac{b^2+c^2-a^2}{2bc}$,
$\implies \tan(A)=\frac{4\triangle}{b^2+c^2-a^2}$
$\implies \tan(A)=\frac{4\triangle}{16\triangle-2bc}$. But the answer is in the form, $\frac{x}{y}$ where $x$ and $y$ are integers.
Any help is appreciated. Thanks in advance.

HINT:
$$16\triangle=(b+c+a)(b+c-a)$$
$$\iff16rs=2s(b+c-a)$$
$$8r=b+c-a$$
Using this and $a=2R\sin A$ etc.,
$$8\cdot4R\prod\sin\dfrac A2=2R\cdot4\cos\dfrac A2\sin\dfrac B2\sin\dfrac C2$$
$$\implies4\sin\dfrac A2=\cos\dfrac A2$$  as $0<B,C<\pi,\sin\dfrac B2\sin\dfrac C2\ne0$
