Proving that, in any $\triangle ABC$, $a(1-\cos^2C)=c(\cos A \cos C + \cos B)$. Question is as follows:

Prove that in any $\triangle ABC$,
$$a(1-\cos^2C)=c(\cos A \cos C + \cos B)$$

I began with Sine rule, i.e. $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k(say)$$
then$$\frac{a}{c}=\frac{k\sin A}{k\sin C}$$
$$=\frac{2\sin A \sin C}{2\sin ^2C}$$
$$=\frac{\cos (A-C) - \cos (A+C)}{2(1-\cos ^2C)}$$
$$\frac{\cos (A-C) + \cos (B)}{2(1-\cos ^2C)}$$
Final step is due to $\cos (A+C)=\cos (180^o-B)=-\cos B$.
My hope was to rearrange $\frac{a}{c}$$\space$ to obtain result but I have been unable to do so.$\space$ I may well have made an error but if anyone can put me on the right track I would be extremely grateful.
Many thanks in advance.
 A: Firstly your denominator should have $(1-\cos^2C)$ not $(1-\cos^2A)$.
Secondly, $$\cos B=\cos(180-(A+C))=-\cos(A+C)=-\cos A\cos C+\sin A\sin C$$
$$\implies\sin A\sin C=\cos B+\cos A\cos C$$
and hence the result
A: We need to prove that:
$$\sin\alpha\sin^2\gamma=\sin\gamma(\cos\alpha\cos\gamma-\cos(\alpha+\gamma))$$ or
$$\sin\alpha\sin\gamma=\cos\alpha\cos\gamma-(\cos\alpha\cos\gamma-\sin\alpha\sin\gamma)$$
Can you end it now?
A: One method using Computer Algebra Systems is to use a
substitution and factorization. In our case, we want to prove

$$ a(1-\cos^2C)=c(\cos A \cos C + \cos B) $$

given $$ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}. $$
The first step is to define substitutions
$$ A = \frac{\log(X)}i, \quad B = \frac{\log(Y)}i,
\quad C = \frac{\log(Z)}i. $$
Rewrite the equation to prove as
$$ \frac{a}{c} = \frac{\sin(A)}{\sin(C)} =
 \frac {\cos A \cos C + \cos B}{\sin^2(C)}. $$
Subtract the last two quotients and use the substitutions to get
$$ \frac{\sin(A)}{\sin(C)} -
 \frac {\cos A \cos C + \cos B}{\sin^2(C)} =
 \frac{ 2 Z (Y + X Z)(1 + X Y Z)}{X Y (1 - Z^2)^2}. $$
Setting this equal to $\,0,\,$ the numerator must equal $\,0.\,$
Since $\,Z\,$ can never equal $\,0,\,$ the numerator will be
$\,0\,$ iff $\,-Y = X Z\,$ or $\,-1 = X Y Z.\,$ The first
case will be true iff $\,180^\circ + B = A + C\,$ and the
second iff $\,180^\circ = A + B + C.\,$ Stated another way,
iff $\,180^\circ = A \pm B + C.\,$ This makes sense since
$\,B\,$ occurs only as $\,\cos B\,$ in the equation to prove.
