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

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(A) Given: $\cos \alpha = \frac{2 \cos \beta - 1}{2 - \cos \beta}$.Using the componendo and dividendo rule:$\frac{1 - \cos \alpha}{1 + \cos \alpha} =

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$\sqrt{3}$

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(A) Given: $\cos \alpha = \frac{2 \cos \beta - 1}{2 - \cos \beta}$.Using the componendo and dividendo rule:$\frac{1 - \cos \alpha}{1 + \cos \alpha} = \frac{(2 - \cos \beta) - (2 \cos \beta - 1)}{(2 - \cos \beta) + (2 \cos \beta - 1)}$$\frac{1 - \cos \alpha}{1 + \cos \alpha} = \frac{2 - \cos \beta - 2 \cos \beta + 1}{2 - \cos \beta + 2 \cos \beta - 1}$$\frac{1 - \cos \alpha}{1 + \cos \alpha} = \frac{3 - 3 \cos \beta}{1 + \cos \beta} = 3 \left( \frac{1 - \cos \beta}{1 + \cos \beta} ight)$Using the half-angle identity $	an^2 	heta = \frac{1 - \cos 2	heta}{1 + \cos 2	heta}$,we get:$	an^2 \frac{\alpha}{2} = 3 	an^2 \frac{\beta}{2}$$\frac{	an^2 \frac{\alpha}{2}}{	an^2 \frac{\beta}{2}} = 3$Since $\frac{1}{	an \frac{\beta}{2}} = \cot \frac{\beta}{2}$,we have:$	an^2 \frac{\alpha}{2} \cot^2 \frac{\beta}{2} = 3$Taking the square root (since $\alpha, \beta \in (0, \pi)$,the tangent values are positive):$	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,where $(0 < \alpha < \pi$ and $0 < \beta < \pi$).

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