Given that,
$$
\begin{aligned}
b \cos (C+\theta)+c \cos (B-\theta) & \\
=b(\cos C & \cos \theta-\sin C \sin \theta) \\
& +c(\cos B \cos \theta+\sin B \sin \theta)
\end{aligned}
$$
$$
\begin{aligned}
=b \cos C \cos \theta+c \cos B \cos \theta-b \sin C & \sin \theta \\
& +c \sin B \sin \theta \\
= & \cos \theta(b \cos C+c \cos B) \\
& -\sin \theta(b \sin C-c \sin B)
\end{aligned}
$$
Since by projection formula and by sine rule
$$
\begin{aligned}
& \frac{b}{\sin B} \frac{c}{\sin C} \Rightarrow b \sin C-c \sin B=0 \\
& b \cos (C+\theta)+c \cos (B-\theta) \\
& =a \cos \theta-\sin \theta \cdot 0 \\
& =a \cos \theta-\sin \theta \cdot 0=a \cos \theta
\end{aligned}
$$