In $\triangle ABC$, $\;$ prove that $\;$ $a \left(\cos B + \cos C\right) = 2 \left(b + c\right) \sin^2 \left(\dfrac{A}{2}\right)$
By cosine rule, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$
By projection formula, $\;$ $b = c \cos A + a \cos C$, $\;$ $c = a \cos B + b \cos A$
$LHS = a \left(\cos B + \cos C\right)$
$= 2 a \cos \left(\dfrac{B + C}{2}\right) \cos \left(\dfrac{B - C}{2}\right)$
$\because \;$ $A$, $B$, $C$ are angles of $\triangle ABC$, $\;$ $A + B + C = \pi$
$\therefore \;$ $\dfrac{B + C}{2} = \dfrac{\pi}{2} - \dfrac{A}{2}$
$\therefore \;$ $\cos \left(\dfrac{B + C}{2}\right) = \cos \left(\dfrac{\pi}{2} - \dfrac{A}{2}\right) = \sin\left(\dfrac{A}{2}\right)$
$\therefore \;$ $LHS = 2 a \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B - C}{2}\right)$ $\;\;\; \cdots \; (1)$
Now, $\;$ $b + c = c \cos A + a \cos C + a \cos B + b \cos A$
i.e. $\;$ $b + c = \left(b + c\right) \cos A + a \left(\cos B + \cos C\right)$
i.e. $\;$ $\left(b + c\right) \left(1 - \cos A\right) = 2 a \cos \left(\dfrac{B + C}{2}\right) \cos \left(\dfrac{B - C}{2}\right)$
i.e. $\;$ $2 \left(b + c\right) \sin^2 \left(\dfrac{A}{2}\right) = 2 a \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B - C}{2}\right)$
i.e. $RHS = 2 a \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B - C}{2}\right)$ $\;\;\; \cdots \; (2)$
$\therefore \;$ From equations $(1)$ and $(2)$, $LHS = RHS$
Hence proved.