If $\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)+1=0$,show that $\alpha-\beta$ or $\beta-\gamma$ or $\gamma-\alpha$ is multiple of $\pi$. This question is from SL Loney.
If $\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)+1=0$,
then show that $\alpha-\beta$ or $\beta-\gamma$ or $\gamma-\alpha$ is a multiple of $\pi$.
My try: Let $\alpha-\beta=A$, $\beta-\gamma=B$, $\gamma-\alpha=C$ so that $A+B+C=0$. So we have to prove that:
If $\cos A +\cos B+\cos C+1=0$, then show that $A,B$ or $C$ is a multiple of $\pi$.
$$2 \cos\frac{A+B}{2}\cos\frac{A-B}{2}+2\cos^2\frac{C}{2}=0\\
2 \cos\frac{C}{2}\cos\frac{A-B}{2}+2\cos^2\frac{C}{2}=0\\
4 \cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2}=0.$$
Either $\cos\frac{A}{2}=0$ or $\cos\frac{B}{2}=0$  or $\cos\frac{C}{2}$.
Either $\frac{A}{2}$=odd multiple of $\frac{\pi}{2}$ or $\frac{B}{2}$=odd multiple of $\frac{\pi}{2}$ or $\frac{C}{2}$=odd multiple of $\frac{\pi}{2}$.
Either $A$=odd multiple of $\pi$ or $B$=odd multiple of $\pi$ or $C$=odd multiple of $\pi$.
But the answer is: 
either $A$=multiple of $\pi$ or $B$=multiple of $\pi$ or $C$=multiple of $\pi$
Is my approach correct? Or is there some other method to prove it.

using your abbreviations, since
$$
\cos A + \cos B + \cos (A+B) + 1 = 0
$$
we have
$$
(1+\cos A)(1+\cos B) = \sin A \sin B
$$
this leads directly to
$$
\cos \frac{A}2 \cos \frac{B}2 (\cos \frac{A}2 \cos \frac{B}2 -\sin \frac{A}2 \sin \frac{B}2)=0
$$
hence either $A=n\pi$ or $B=n\pi$ or 
$$
\cos \frac{A}2 \cos \frac{B}2 -\sin \frac{A}2 \sin \frac{B}2=0
$$
which gives
$$
\tan \frac{A}2 = \cot \frac{B}2
$$
or
$$
A+B = (2n+1)\pi
$$
