In $\triangle ABC$, prove that $2 \left[a \sin^2 \left(\dfrac{C}{2}\right) + c \sin^2 \left(\dfrac{A}{2}\right) \right] = c + a - b$
By cosine rule,
$\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$
$\begin{aligned}
LHS & = 2 \left[a \sin^2 \left(\dfrac{C}{2}\right) + c \sin^2 \left(\dfrac{A}{2}\right) \right] \\\\
& \left[\text{Note: } \sin^2 \left(\dfrac{\theta}{2}\right) = \dfrac{1 - \cos \theta}{2}\right] \\\\
& = 2 \left[a \left(\dfrac{1 - \cos C}{2}\right) + c \left(\dfrac{1 - \cos A}{2}\right)\right] \\\\
& = a \left[1 - \left(\dfrac{a^2 + b^2 - c^2}{2 a b}\right)\right] + c \left[1 - \left(\dfrac{b^2 + c^2 - a^2}{2 b c}\right)\right] \;\;\; \left[\text{by cosine rule}\right] \\\\
& = a \left(\dfrac{2 a b - a^2 - b^2 + c^2}{2 a b}\right) + c \left(\dfrac{2 b c - b^2 - c^2 + a^2}{2 b c}\right) \\\\
& = \dfrac{1}{2 b} \left(2 a b - a^2 - b^2 + c^2 + 2 b c - b^2 - c^2 + a^2\right) \\\\
& = \dfrac{1}{2 b} \left(2 a b + 2 b c - 2 b^2\right) \\\\
& = a + c - b = RHS
\end{aligned}$
Hence proved.