Mathematics - Properties of Triangles Question with Solution | TestHub

Given, $\cos B+\cos C=4{\sin }^2\frac{A}{2}$ $$\begin{array}{rl}& \Rightarrow 2\cos \left(\frac{B+C}{2}\right)\cos \left(\frac{B-C}{2}\right)=4{\sin }^2\frac{A}{2}\\ & \Rightarrow 2\sin \frac{A}{2}\left[\cos \left(\frac{B-C}{2}\right)-2\sin \frac{A}{2}\right]=0\\ & \Rightarrow \cos \left(\frac{B-C}{2}\right)-2\cos \left(\frac{B+C}{2}\right)=0\\ & \Rightarrow -\cos \frac{B}{2}\cos \frac{C}{2}+3\sin \frac{B}{2}\sin \frac{C}{2}=0\\ & \Rightarrow \tan \frac{B}{2}\tan \frac{C}{2}=\frac{1}{3}\\ & \Rightarrow \sqrt{\frac{(s-a)(s-c)}{s(s-b)}\cdot \frac{(s-b)(s-a)}{s(s-c)}}=\frac{1}{3}\\ & \Rightarrow \frac{s-a}{s}=\frac{1}{3}\Rightarrow 2s=3a\\ & \Rightarrow b+c=2a\end{array}$$
$\therefore$ Locus of $A$ is an ellipse.