If in a triangle $ABC, a^2+2bc-\left(b^2+c^2\right)=ab\sin \frac{C}{2}\cos \frac{C}{2}$, then $\cot (B+C)=$

Given, $a^2+2bc-\left(b^2+c^2\right)=ab\sin \frac{C}{2}\cos \frac{C}{2}$

$\Rightarrow 2bc-\left(b^2+c^2-a^2\right)=ab\sin \frac{C}{2}\cos \frac{C}{2}$

$\Rightarrow 2bc-2bc\cos A=ab\sin \frac{C}{2}\cos \frac{C}{2}$

$\Rightarrow 2c\left(1-\cos A\right)=a\sin \frac{C}{2}\cos \frac{C}{2}$

$\Rightarrow 4c.{\sin }^2\frac{A}{2}=a\sin \frac{C}{2}\cos \frac{C}{2}$

$\Rightarrow 4c.\frac{\left(s-b\right)\left(s-c\right)}{bc}=a\sqrt{\frac{\left(s-a\right)\left(s-b\right)}{ab}}\sqrt{\frac{s\left(s-c\right)}{ab}}$

$\Rightarrow \sqrt{\frac{\left(s-b\right)\left(s-c\right)}{s\left(s-a\right)}}=\frac{1}{4}$

$\Rightarrow \tan \left(\frac{A}{2}\right)=\frac{1}{4}$

$\Rightarrow \tan \left(A\right)=\frac{2\tan \left({\displaystyle \frac{A}{2}}\right)}{1-{\tan }^2\left(\frac{A}{2}\right)}=\frac{{\displaystyle \frac{1}{2}}}{1-{\displaystyle \frac{1}{16}}}=\frac{8}{15}$

$\Rightarrow \tan \left(\mathrm{\pi }-\left(B+C\right)\right)=\frac{8}{15}$

$\Rightarrow -\tan \left(B+C\right)=\frac{8}{15}$

$\Rightarrow \cot \left(B+C\right)=\frac{-15}{8}$