If sides $a$, $b$, $c$ of $\triangle ABC$ are in arithmetic progression, then $3\tan\frac{A}{2}\tan\frac {C}{2}=1$
Expand your last line: $$2\left(\cos\frac A2\cos\frac C2 - \sin\frac A2\sin\frac C2\right)=\left(\cos\frac A2\cos\frac C2 +\sin\frac A2\sin\frac C2\right)$$ and your result is immediate after a cancellation.
Hint:
$$\dfrac21=\dfrac{\cos\dfrac{A-C}2}{\cos\dfrac{A+C}2}$$
Apply Componendo and Dividendo
$$\dfrac{2+1}{2-1}=?$$