Properties of Triangle - Trigo Problem : In $\triangle $ABC prove that $a\cos(C+\theta) +c\cos(A-\theta) = b\cos\theta$ Problem : 
In $\triangle $ABC prove that $a\cos(C+\theta) +\cos(A-\theta) = b\cos\theta$
My approach : 
Using $\cos(A+B) =\cos A\cos B -\sin A\sin B and \cos(A-B) = \cos A\cos B +\sin A\sin B$, we get: 
\begin{equation}
a(\cos C\cos\theta -\sin C\sin\theta) +c(\cos A\cos\theta +\sin A\sin\theta)\quad\\=\cos\theta(a\cos C+c\cos A) +\sin\theta(c\sin A -a\sin C)\qquad(\text{i})
\end{equation}
By using projection formula which states: $b=a\cos C +c \cos A $
(i) will become : 
$$b\cos\theta +\sin\theta (c\sin A -a\sin C)$$ How do I proceed from here? 

Law of sines states that
$$\frac{\sin{A}}{a} = \frac{\sin{C}}{c}$$
which means that $c \sin{A} - a \sin{C} = 0$.  The desired result follows.
