$Let α and β be nonzero real numbers such that 2 (\cos β - \cos α) + \cos α \cos β = 1 . Then which of the$ $f o l l o w i n g i s / a r e t r u e ?$

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$\tan (\frac{α}{2}) + \sqrt{3} \tan (\frac{β}{2}) = 0$

$\sqrt{3} \tan (\frac{α}{2}) - \tan (\frac{β}{2}) = 0$

$\tan (\frac{α}{2}) - \sqrt{3} \tan (\frac{β}{2}) = 0$

$\sqrt{3} \tan (\frac{α}{2}) + \tan (\frac{β}{2}) = 0$

answer is A, C.

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Detailed Solution

$\begin{array}{l} 2 (\cos β - \cos α) + \cos α \cos β = 1 \\ \Rightarrow \cos β = \frac{1 + 2 \cos α}{2 + \cos α} \\ \Rightarrow \frac{1 - y^{2}}{1 + y^{2}} = \frac{1 + 2 (\frac{1 - x^{2}}{1 + x^{2}})}{2 + \frac{1 - x^{2}}{1 + x^{2}}} (x = \tan α / 2, y = \tan β / 2) \\ \Rightarrow \frac{1 - y^{2}}{1 + y^{2}} = \frac{3 - x^{2}}{3 + x^{2}} \Rightarrow 3 y^{2} = x^{2} \\ (\sqrt{3} \tan \frac{β}{2} + \tan \frac{α}{2}) (\sqrt{3} \tan \frac{β}{2} - \tan \frac{α}{2}) = 0 \end{array}$