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.
Which means that $c \sin(B+C)=a \sin(A+B)$

$$c\sin(B+C)-a\sin(A+B)=O$$