<s>[INST] If $2 (\cos \alpha-\cos\beta)+\cos\alpha\cos\beta=1$, then $p\tan\frac{\alpha}{2}+q\tan\frac{\beta}{2}=0$ for which choices of $p$ and $q$? Let $\alpha$ and $\beta$ be nonzero real numbers such that $2 (\cos \alpha-\cos\beta)+\cos\alpha\cos\beta=1$. Then which of the following is/are true ?
a)  $\sqrt3\tan(\frac\alpha2)+\tan(\frac\beta2)=0$ 
b)  $\sqrt3\tan(\frac\alpha2)-\tan(\frac\beta2)=0$
c)  $\tan(\frac\alpha2)+\sqrt3\tan(\frac\beta2)=0$
d)  $\tan(\frac\alpha2)-\sqrt3\tan(\frac\beta2)=0$
I solved the equation by converting all cosines to half angles and the answers options c) and d), seems pretty straightforward from there. But, when the question was asked in an engineering entrance exam (JEE ADVANCED  2017), the question was given as BONUS that would mean either the question was not complete, was wrong or might be some other mistake in understanding the question. I do not understand what was the problem. Can someone help me point out the ambiguity or mistake in the question or give a solution that differs from mine.
 [/INST] Given $2(\cos \alpha-\cos \beta)+\cos \alpha\cos \beta = 1$
So $\displaystyle \cos \alpha(\cos \beta+2)=1+2\cos \beta \Rightarrow \cos \alpha =\frac{2\cos \beta+1}{2+\cos \beta}$
So $\displaystyle \frac{1}{\cos \alpha} = \frac{2+\cos \beta}{2\cos \beta +1}$
Using Componendo Dividedno, We have 
$\displaystyle \frac{1+\cos \alpha}{1-\cos \alpha} = \frac{2+\cos \beta+2\cos \beta+1}{(2+\cos \beta)-(2\cos \beta +1)} = \frac{3(\cos \beta+1)}{1-\cos \beta}$
So $\displaystyle \frac{2\sin^2\frac{\alpha}{2}}{2\cos^2 \frac{\alpha}{2}} = 3\cdot \frac{2\sin^2\frac{\beta}{2}}{2\cos^2 \frac{\beta}{2}}$
So we have $$\tan \frac{\alpha}{2} = \pm\sqrt{3}\tan \frac{\beta}{2}.$$
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