If cos⁡x−cos⁡αcos⁡x−cos⁡β=sin2αcos⁡βsin2⁡βcos⁡α, then

tan ⁡ x 2 = tan ⁡ α 2 tan ⁡ β 2

If cos⁡x−cos⁡αcos⁡x−cos⁡β=sin2αcos⁡βsin2⁡βcos⁡α, then

tan ⁡ x 2 = tan ⁡ α 2 tan ⁡ β 2

cos ⁡ x = 2 cos α + cos β 1 + cos α cos β

cos ⁡ x = cos ⁡ a + cos ⁡ β 1 − cos α cos ⁡ β

tan ⁡ x 2 = tan ⁡ α 2 tan ⁡ β 2

tan ⁡ x 2 = − 2 tan ⁡ α 2 tan ⁡ β 2

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$If \frac{\cos x - \cos α}{\cos x - \cos β} = \frac{\sin^{2} α \cos β}{\sin^{2} β \cos α}, then$

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$\cos x = \frac{2 \cos α + \cos β}{1 + \cos α \cos β}$

$\cos x = \frac{\cos a + \cos β}{1 - \cos α \cos β}$

$\tan \frac{x}{2} = \tan \frac{α}{2} \tan \frac{β}{2}$

$\tan \frac{x}{2} = - 2 \tan \frac{α}{2} \tan \frac{β}{2}$

answer is C.

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Detailed Solution

From the given results

$\begin{array}{l} \cos x \sin^{2} β \cos α - \cos^{2} α \sin^{2} β \\ = \sin^{2} α \cos β \cos x - \sin^{2} α \cos^{2} β \\ \Rightarrow \cos x [\cos α \sin^{2} β - \sin^{2} α \cos β] \\ = \cos^{2} α \sin^{2} β - \sin^{2} α \cos^{2} β \end{array}$

$\begin{array}{l} \Rightarrow \cos x = \frac{\cos^{2} α [1 - \cos^{2} β] - (1 - \cos^{2} α) \cos^{2} β}{\cos α [1 - \cos^{2} β] - [1 - \cos^{2} α] \cos β} \\ = \frac{\cos^{2} α - \cos^{2} β}{(\cos α - \cos β) (\cos α \cos β + 1)} = \frac{\cos α + \cos β}{1 + \cos α \cos β} \\ \Rightarrow \frac{\cos x}{1} = \frac{\cos α + \cos β}{1 + \cos α \cos β} \\ \Rightarrow \frac{1 - \cos x}{1 + \cos x} = \frac{1 + \cos α \cos β - \cos α - \cos β}{1 + \cos α \cos β + \cos α + \cos β} \\ = \frac{(1 - \cos α) (1 - \cos β)}{(1 + \cos α) (1 + \cos β)} \end{array}$

$\begin{array}{l} \Rightarrow \frac{1 - \cos x}{1 + \cos x} = \frac{1 + \cos α \cos β - \cos α - \cos β}{1 + \cos α \cos β + \cos α + \cos β} \\ = \frac{(1 - \cos α) (1 - \cos β)}{(1 + \cos α) (1 + \cos β)} \\ \Rightarrow \tan^{2} \frac{x}{2} = \tan^{2} \frac{α}{2} \cdot \tan^{2} \frac{β}{2} \Rightarrow \tan \frac{x}{2} = \pm \tan \frac{α}{2} \tan \frac{β}{2} \\ \Rightarrow \tan \frac{x}{2} = \pm \tan \frac{α}{2} \tan \frac{β}{2} \end{array}$