Đk $0^o<A<180^o\Rightarrow |\cos A| <1\Rightarrow \begin{cases}\cos A+1 >0 \\ \cos A-1<0 \end{cases}$
Đặt $M= \frac{2(b^2+c^2-a^2)}{(b+c)^2-a^2}\Rightarrow M=\frac{4bc .\cos A}{2bc( \cos A-1)}=\frac{2\cos A}{ \cos A-1}$
Và $M+1=\frac{3b^2+3c^2-3a^2+2bc}{(b+c)^2-a^2}$
$pt\Leftrightarrow 3[(M+1)^4-1]+4 [\tan^6 \frac A2-1]=0$
$\Leftrightarrow 3[(M+1)^2-1][(M+1)^2+1]+4(\tan ^2 \frac A2-1)( \tan^4 \frac A2+\tan ^2 \frac A2+1)=0$
$\Leftrightarrow 3M(M+2)[(M+1)^2+1]+4(\frac{1-cos A}{1+\cos A}-1)( \tan^4 \frac A2+\tan ^2 \frac A2+1)=0$
$\Leftrightarrow 3.\frac{2 \cos A}{\cos A-1}.(M+2)[(M+1)^2+1]+4.\frac{-2cosA}{\cos A+1}.( \tan^4 \frac A2+\tan ^2 \frac A2+1)=0$
$\Leftrightarrow \cos A. \left[ \frac{6}{\cos A-1 }.(M+2)(M^2+2M+2)- \frac{8}{\cos A+1}(\tan^4 \frac A2+\tan ^2 \frac A2+1)\right]=0$
Trong ngoặc vuông $<0\Rightarrow \cos A=0\Rightarrow A=90^o$