If cos x = frac{2 cos y - 1}{2 - cos y} where | vedclass

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(B) Given $\cos x = \frac{2 \cos y - 1}{2 - \cos y}$.Applying the componendo and dividendo rule:$\frac{1 - \cos x}{1 + \cos x} = \frac{(2 - \cos y) -

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$\sqrt{3}$

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(B) Given $\cos x = \frac{2 \cos y - 1}{2 - \cos y}$.Applying the componendo and dividendo rule:$\frac{1 - \cos x}{1 + \cos x} = \frac{(2 - \cos y) - (2 \cos y - 1)}{(2 - \cos y) + (2 \cos y - 1)}$$\frac{2 \sin^2(x/2)}{2 \cos^2(x/2)} = \frac{3 - 3 \cos y}{1 + \cos y}$$	an^2(x/2) = \frac{3(1 - \cos y)}{1 + \cos y} = \frac{3(2 \sin^2(y/2))}{2 \cos^2(y/2)}$$	an^2(x/2) = 3 	an^2(y/2)$Since $x, y \in (0, \pi)$,$x/2, y/2 \in (0, \pi/2)$,so $	an(x/2)$ and $	an(y/2)$ are positive.Taking the square root,we get $	an(x/2) = \sqrt{3} 	an(y/2)$.Therefore,$	an(x/2) \cot(y/2) = \sqrt{3}$.

If $\cos x = \frac{2 \cos y - 1}{2 - \cos y}$ where $x, y \in (0, \pi)$,then $\tan(x/2) \cot(y/2) =$

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