In a $∆ABC$ with fixed base BC, the vertex A moves such that $\cos B+\cos C=4{\sin }^2\frac{A}{2}.$ If $a, b and c$ denote lengths of the sides of triangle opposite to the angles A, B and C respectively, then

Given, $\cos B+\cos C=4{\sin }^2\frac{A}{2}$

$\Rightarrow 2\cos \left(\frac{B+C}{2}\right)\cos \left(\frac{B-C}{2}\right)=4{\sin }^2\frac{A}{2}$

$\begin{array}{r}\Rightarrow 2\sin \frac{A}{2}\left[\cos \left(\frac{B-C}{2}\right)-2\sin \frac{A}{2}\right]=0\\ \Rightarrow \cos \left(\frac{B-C}{2}\right)-2\cos \left(\frac{B+C}{2}\right)=0\end{array}$

as  $\sin \frac{A}{2}\ne 0$

$\begin{array}{l}\Rightarrow -\cos \frac{B}{2}\cos \frac{C}{2}+3\sin \frac{B}{2}\sin \frac{C}{2}=0\\ \Rightarrow \tan \frac{B}{2}\tan \frac{C}{2}=\frac{1}{3}\end{array}$

$\Rightarrow \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}\cdot \sqrt{\frac{(s-b)(s-a)}{s(s-c)}}=\frac{1}{3}\Rightarrow \frac{s-a}{s}=\frac{1}{3}$

$\begin{array}{l}\Rightarrow 2s=3a\\ \therefore b+c=2a [option (b\left)\right] \end{array}$

$\therefore$ Locus of A is an ellipse.  [option (c)]