In $\triangle ABC$, prove that $\dfrac{1 + \cos A \cos \left(B - C\right)}{1 + \cos C \cos \left(A - B\right)} = \dfrac{b^2 + c^2}{b^2 + a^2}$
$\left\{\text{Note: } \cos \alpha \cos \beta = \dfrac{1}{2} \left[\cos \left(\alpha + \beta\right) + \cos \left(\alpha - \beta\right)\right] \right\}$
$\begin{aligned}
LHS & = \dfrac{1 + \cos A \cos \left(B - C\right)}{1 + \cos C \cos \left(A - B\right)} \\\\
& = \dfrac{1 + \dfrac{1}{2} \left[\cos \left(A + B - C\right) + \cos \left(A - B + C\right)\right]}{1 + \dfrac{1}{2} \left[\cos \left(C + A - B\right) + \cos \left(C - A + B\right)\right]} \\\\
& = \dfrac{2 + \cos \left(A + B - C\right) + \cos \left(A - B + C\right)}{2 + \cos \left(C + A - B\right) + \cos \left(C - A + B\right)} \;\;\; \cdots \; (1)
\end{aligned}$
In $\;$ $\triangle ABC$, $\;$ $A + B + C = \pi$
$\therefore \;$ $A + B = \pi - C$ $\;\; \cdots \; (2a)$,
$A + C = \pi - B$ $\;\; \cdots \; (2b)$,
$B + C = \pi - A$ $\;\; \cdots \; (2c)$
$\therefore \;$ In view of equations $(2a)$, $(2b)$ and $(2c)$ equation $(1)$ becomes,
$\begin{aligned}
LHS & = \dfrac{2 + \cos \left(\pi - C - C\right) + \cos \left(\pi - B - B\right)}{2 + \cos \left(\pi - B - B\right) + \cos \left(\pi - A - A\right)} \\\\
& = \dfrac{2 + \cos \left(\pi - 2 C\right) + \cos \left(\pi - 2 B\right)}{2 + \cos \left(\pi - 2 B\right) + \cos \left(\pi - 2 A\right)} \\\\
& \left[\text{Note: } \cos \left(\pi - \theta\right) = - \cos \theta\right] \\\\
& = \dfrac{2 - \cos \left(2 C\right) - \cos \left(2 B\right)}{2 - \cos \left(2 B\right) - \cos \left(2 A\right)} \\\\
& \left[\text{Note: } \cos \left(2 \theta\right) = 1 - 2 \sin^2 \theta\right] \\\\
& = \dfrac{2 - \left(1 - 2 \sin^2 C\right) - \left(1 - 2 \sin^2 B\right)}{2 - \left(1 - 2 \sin^2 B\right) - \left(1 - 2 \sin^2 A\right)} \\\\
& = \dfrac{2 - 1 + 2 \sin^2 C - 1 + 2 \sin^2 B}{2 - 1 + 2 \sin^2 B - 1 + 2 \sin^2 A} \\\\
& = \dfrac{\sin^2 C + \sin^2 B}{\sin^2 B + \sin^2 A} \;\;\; \cdots \; (3)
\end{aligned}$
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
$\implies$ $\sin A = \dfrac{a}{2 R}$ $\;\; \cdots \; (4a)$, $\;$ $\sin B = \dfrac{b}{2 R}$ $\;\; \cdots \; (4b)$, $\;$ $\sin C = \dfrac{c}{2 R}$ $\;\; \cdots \; (4c)$
$\therefore \;$ In view of equations $(4a)$, $(4b)$ and $(4c)$ equation $(3)$ becomes,
$\begin{aligned}
LHS & = \dfrac{\dfrac{c^2}{4 R^2} + \dfrac{b^2}{4 R^2}}{\dfrac{b^2}{4 R^2} + \dfrac{a^2}{4 R^2}} \\\\
& = \dfrac{c^2 + b^2}{b^2 + a^2} \\\\
& = RHS
\end{aligned}$
Hence proved.