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

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(A) Given,$\cos 	heta = \frac{\cos \alpha - \cos \beta}{1 - \cos \alpha \cos \beta}$.Applying componendo and dividendo,we get:$\frac{1 + \cos 	heta}

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$\cot \frac{\beta}{2} 	an \frac{\alpha}{2}$

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(A) Given,$\cos 	heta = \frac{\cos \alpha - \cos \beta}{1 - \cos \alpha \cos \beta}$.Applying componendo and dividendo,we get:$\frac{1 + \cos 	heta}{1 - \cos 	heta} = \frac{1 + \frac{\cos \alpha - \cos \beta}{1 - \cos \alpha \cos \beta}}{1 - \frac{\cos \alpha - \cos \beta}{1 - \cos \alpha \cos \beta}}$$\frac{1 + \cos 	heta}{1 - \cos 	heta} = \frac{1 - \cos \alpha \cos \beta + \cos \alpha - \cos \beta}{1 - \cos \alpha \cos \beta - \cos \alpha + \cos \beta}$$\frac{1 + \cos 	heta}{1 - \cos 	heta} = \frac{(1 + \cos \alpha)(1 - \cos \beta)}{(1 - \cos \alpha)(1 + \cos \beta)}$Using half-angle formulas,$\frac{1 + \cos x}{1 - \cos x} = \cot^2 \frac{x}{2}$ and $\frac{1 - \cos x}{1 + \cos x} = 	an^2 \frac{x}{2}$:$\cot^2 \frac{	heta}{2} = \cot^2 \frac{\alpha}{2} 	an^2 \frac{\beta}{2}$Taking the reciprocal on both sides:$	an^2 \frac{	heta}{2} = 	an^2 \frac{\alpha}{2} \cot^2 \frac{\beta}{2}$Taking the square root:$	an \frac{	heta}{2} = 	an \frac{\alpha}{2} \cot \frac{\beta}{2}$

If $\cos \theta = \frac{\cos \alpha - \cos \beta}{1 - \cos \alpha \cos \beta}$,then one of the values of $\tan \frac{\theta}{2}$ is

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