Answer: $\mathbf A$
We have:
$\mathrm {\angle A + \angle B + \angle C } = \pi \;\text{(By angle sum property of $\Delta$)}$
$\Rightarrow \mathrm{\angle A + \dfrac{2\pi}{3} + \angle C}=\dfrac{\pi}{3}$
Now, We know that:
$\mathrm{\cos A + \cos C = 2\cos\frac{A+C}{2}.\cos\frac{A-C}{2}}$
$\Rightarrow \mathrm{\cos A + \cos C = 2\cos\frac{\pi}{6}.\cos\frac{A-C}{2}} = 2\times \frac{\sqrt3}{2}\cos\frac{A-C}{2} = \sqrt3\cos\frac{A-C}{2}$
$\Rightarrow \mathrm{\cos A + \cos C=\sqrt3\cos\frac{A-C}{2}} $
We know that:
$-1 \le\cos \theta \le1, \;\forall \theta$
$\mathrm{\Rightarrow -1\le \cos\big (\frac{A-C}{2} \big )\le1}$
$\mathrm{\Rightarrow -\sqrt3\le \sqrt 3\cos\big (\frac{A-C}{2} \big )\le \sqrt3}$
$\Rightarrow \text {Range} = [-\sqrt3, \;\sqrt3]$
$\therefore \mathbf A$ is the right answer.
