If $\cos x - \sin α \cot β \sin x = \cos α$ , then the value of $\tan (x / 2)$ is

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$- \tan (α / 2) \cot (β / 2)$

$\tan (α / 2) \tan (β / 2)$

$- \cot (α / 2) \tan (β / 2)$

$\cot (α / 2) \cot (β / 2)$

answer is A, B.

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Detailed Solution

$\begin{array}{l} \cos x - \sin αcot βsin x = \cos α \\ \Rightarrow \frac{1 - \tan^{2} (x / 2)}{1 + \tan^{2} (x / 2)} - \sin αcot β \frac{2 \tan (x / 2)}{1 + \tan^{2} (x / 2)} = \cos α \\ \Rightarrow \tan^{2} \frac{x}{2} (1 + \cos α) + 2 \sin αcot βtan \frac{x}{2} - (1 - \cos α) = 0 \\ \Rightarrow \tan^{2} \frac{x}{2} + \frac{2 \sin αcot β}{1 + \cos α} \tan \frac{x}{2} - \frac{1 - \cos α}{1 + \cos α} = 0 \\ \Rightarrow \tan^{2} \frac{x}{2} + 2 \tan \frac{α}{2} \cot βtan \frac{x}{2} - \tan^{2} \frac{α}{2} = 0 \\ \Rightarrow \tan^{2} \frac{x}{2} + 2 \tan \frac{α}{2} \cdot \frac{1}{2} (\cot \frac{β}{2} - \tan \frac{β}{2}) \tan \frac{x}{2} - \tan^{2} \frac{α}{2} = 0 \end{array}$

$\begin{array}{l} \Rightarrow (\tan \frac{x}{2} + \cot \frac{β}{2} \tan \frac{α}{2}) (\tan \frac{x}{2} - \tan \frac{β}{2} \tan \frac{α}{2}) = 0 \\ \Rightarrow \tan (\frac{x}{2}) = - \tan (\frac{α}{2}) \cot (\frac{β}{2}) \\ or \tan (\frac{x}{2}) = \tan (\frac{α}{2}) \tan (\frac{β}{2}) \end{array}$