If tan beta = cos theta tan alpha , then {tan | vedclass

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(C) Given $	an \beta = \cos 	heta 	an \alpha$,we have $\cos 	heta = \frac{	an \beta}{	an \alpha} = \frac{\sin \beta \cos \alpha}{\cos \beta \sin

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$\frac{\sin (\alpha - \beta )}{\sin (\alpha + \beta )}$

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(C) Given $	an \beta = \cos 	heta 	an \alpha$,we have $\cos 	heta = \frac{	an \beta}{	an \alpha} = \frac{\sin \beta \cos \alpha}{\cos \beta \sin \alpha}$.Using the formula ${	an ^2}\frac{	heta }{2} = \frac{1 - \cos 	heta}{1 + \cos 	heta}$,we substitute $\cos 	heta$:${	an ^2}\frac{	heta }{2} = \frac{1 - \frac{\sin \beta \cos \alpha}{\cos \beta \sin \alpha}}{1 + \frac{\sin \beta \cos \alpha}{\cos \beta \sin \alpha}}$$= \frac{\sin \alpha \cos \beta - \cos \alpha \sin \beta}{\sin \alpha \cos \beta + \cos \alpha \sin \beta}$$= \frac{\sin (\alpha - \beta )}{\sin (\alpha + \beta )}$.

If $\tan \beta = \cos \theta \tan \alpha ,$ then ${\tan ^2}\frac{\theta }{2} = $

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