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.

We have $16\triangle=(b+c+a)(b+c-a)$
$\iff16r\cdot s=2s\cdot2(s-a)$ where $r$ is the in-radius & $2s=a+b+c$
Using this, $\tan\dfrac A2=\dfrac r{s-a}=\dfrac14$
Finally use $\tan2x$ formula
