If cos alpha = frac{l cos beta + m}{l + m cos | vedclass

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Question

(A) Given $\cos \alpha = \frac{l \cos \beta + m}{l + m \cos \beta}$.Using the formula $	an^2 \frac{	heta}{2} = \frac{1 - \cos 	heta}{1 + \cos 	het

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$\frac{l - m}{l + m}$

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(A) Given $\cos \alpha = \frac{l \cos \beta + m}{l + m \cos \beta}$.Using the formula $	an^2 \frac{	heta}{2} = \frac{1 - \cos 	heta}{1 + \cos 	heta}$,we have:$	an^2 \frac{\alpha}{2} = \frac{1 - \cos \alpha}{1 + \cos \alpha} = \frac{1 - \frac{l \cos \beta + m}{l + m \cos \beta}}{1 + \frac{l \cos \beta + m}{l + m \cos \beta}}$$= \frac{l + m \cos \beta - l \cos \beta - m}{l + m \cos \beta + l \cos \beta + m} = \frac{l(1 - \cos \beta) - m(1 - \cos \beta)}{l(1 + \cos \beta) + m(1 + \cos \beta)}$$= \frac{(l - m)(1 - \cos \beta)}{(l + m)(1 + \cos \beta)} = \frac{l - m}{l + m} \cdot 	an^2 \frac{\beta}{2}$.Therefore,$\left(\frac{	an \frac{\alpha}{2}}{	an \frac{\beta}{2}}ight)^2 = \frac{l - m}{l + m}$.

If $\cos \alpha = \frac{l \cos \beta + m}{l + m \cos \beta}$,then $\left(\frac{\tan \frac{\alpha}{2}}{\tan \frac{\beta}{2}}\right)^2 = $

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