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.
 A: $$b^2+c^2-a^2=16\Delta - 2bc \Rightarrow \frac{b^2+c^2-a^2}{2bc}=\frac{8\Delta}{bc}-1 \Rightarrow \cos A=4\sin A-1$$
$$\Rightarrow 2\cos^2\frac{A}{2}-1=8\sin\frac{A}{2}\cos\frac{A}{2}-1 \Rightarrow \tan\frac{A}{2}=\frac{1}{4}$$
Hence,
$$\tan A=\frac{2\cdot \frac{1}{4}}{1-\frac{1}{4^2}}=\boxed{\dfrac{8}{15}}$$
A: 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 $0<B,C<\pi,\sin\dfrac B2\sin\dfrac C2\ne0$
A: You have: $\dfrac{\sin A}{\cos A} =\tan A = \dfrac{4\triangle}{16\triangle - 2bc} = \dfrac{2bc\sin A}{8bc\sin A-2bc} = \dfrac{\sin A}{4\sin A-1} \Rightarrow \cos A = 4\sin A -1 \Rightarrow \cos^2A = \left(1-4\sin A\right)^2 = 1 - 8\sin A + 16\sin^2A \Rightarrow 1 - \sin^2A = 1 - 8\sin A + 16\sin^2A \Rightarrow \sin A\left(8 - 17\sin A\right) = 0 \Rightarrow \sin A = 0 \text{ or} \text{ }\sin A = \dfrac{8}{17}$. But $\sin A > 0$, so $\sin A = \dfrac{8}{17} \Rightarrow \cos A = 4\sin A -1  = 4\cdot \dfrac{8}{17} - 1 = \dfrac{15}{17} \Rightarrow \tan A = \dfrac{\dfrac{8}{17}}{\dfrac{15}{17}} = \dfrac{8}{15}$.
A: 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
