If cos alpha = frac{2 cos beta - 1}{2 - cos b | vedclass

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(A) Given $\cos \alpha = \frac{2 \cos \beta - 1}{2 - \cos \beta}$.Applying Componendo and Dividendo:$\frac{1 - \cos \alpha}{1 + \cos \alpha} = \frac{(

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$\sqrt{3}$

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(A) Given $\cos \alpha = \frac{2 \cos \beta - 1}{2 - \cos \beta}$.Applying Componendo and Dividendo:$\frac{1 - \cos \alpha}{1 + \cos \alpha} = \frac{(2 - \cos \beta) - (2 \cos \beta - 1)}{(2 - \cos \beta) + (2 \cos \beta - 1)}$$\frac{2 \sin^2 \frac{\alpha}{2}}{2 \cos^2 \frac{\alpha}{2}} = \frac{3 - 3 \cos \beta}{1 + \cos \beta}$$	an^2 \frac{\alpha}{2} = \frac{3(1 - \cos \beta)}{1 + \cos \beta}$$	an^2 \frac{\alpha}{2} = 3 	an^2 \frac{\beta}{2}$$	an^2 \frac{\alpha}{2} \cot^2 \frac{\beta}{2} = 3$Taking the square root,$	an \frac{\alpha}{2} \cot \frac{\beta}{2} = \sqrt{3}$.

If $\cos \alpha = \frac{2 \cos \beta - 1}{2 - \cos \beta}$,then the value of $\tan \frac{\alpha}{2} \cot \frac{\beta}{2}$ is equal to,where $(0 < \alpha < \pi$ and $0 < \beta < \pi)$.

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