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

From the given results cos⁡xsin2⁡βcos⁡α−cos2⁡αsin2⁡β=sin2⁡αcos⁡βcos⁡x−sin2⁡αcos2⁡β⇒ cos⁡xcos⁡αsin2⁡β−sin2⁡αcos⁡β=cos2⁡αsin2⁡β−sin2⁡αcos2⁡β⇒ cos⁡x=cos2⁡α1−cos2⁡β−1−cos2⁡αcos2⁡βcos⁡α1−cos2⁡β−1−cos2⁡αcos⁡β=cos2⁡α−cos2⁡β(cos⁡α−cos⁡β)(cos⁡αcos⁡β+1)=cos⁡α+cos⁡β1+cos⁡αcos⁡β⇒cos⁡x1=cos⁡α+cos⁡β1+cos⁡αcos⁡β⇒1−cos⁡x1+cos⁡x=1+cos⁡αcos⁡β−cos⁡α−cos⁡β1+cos⁡αcos⁡β+cos⁡α+cos⁡β=(1−cos⁡α)(1−cos⁡β)(1+cos⁡α)(1+cos⁡β)⇒1−cos⁡x1+cos⁡x=1+cos⁡αcos⁡β−cos⁡α−cos⁡β1+cos⁡αcos⁡β+cos⁡α+cos⁡β=(1−cos⁡α)(1−cos⁡β)(1+cos⁡α)(1+cos⁡β)⇒tan2⁡x2=tan2⁡α2⋅tan2⁡β2⇒tan⁡x2=±tan⁡α2tan⁡β2⇒tan⁡x2=±tan⁡α2tan⁡β2

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

From the given results cos⁡xsin2⁡βcos⁡α−cos2⁡αsin2⁡β=sin2⁡αcos⁡βcos⁡x−sin2⁡αcos2⁡β⇒ cos⁡xcos⁡αsin2⁡β−sin2⁡αcos⁡β=cos2⁡αsin2⁡β−sin2⁡αcos2⁡β⇒ cos⁡x=cos2⁡α1−cos2⁡β−1−cos2⁡αcos2⁡βcos⁡α1−cos2⁡β−1−cos2⁡αcos⁡β=cos2⁡α−cos2⁡β(cos⁡α−cos⁡β)(cos⁡αcos⁡β+1)=cos⁡α+cos⁡β1+cos⁡αcos⁡β⇒cos⁡x1=cos⁡α+cos⁡β1+cos⁡αcos⁡β⇒1−cos⁡x1+cos⁡x=1+cos⁡αcos⁡β−cos⁡α−cos⁡β1+cos⁡αcos⁡β+cos⁡α+cos⁡β=(1−cos⁡α)(1−cos⁡β)(1+cos⁡α)(1+cos⁡β)⇒1−cos⁡x1+cos⁡x=1+cos⁡αcos⁡β−cos⁡α−cos⁡β1+cos⁡αcos⁡β+cos⁡α+cos⁡β=(1−cos⁡α)(1−cos⁡β)(1+cos⁡α)(1+cos⁡β)⇒tan2⁡x2=tan2⁡α2⋅tan2⁡β2⇒tan⁡x2=±tan⁡α2tan⁡β2⇒tan⁡x2=±tan⁡α2tan⁡β2

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

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

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

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

answer is C.

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Detailed Solution

From the given results

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

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

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