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

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(C) Given that $	an \beta = \cos 	heta 	an \alpha$,we have $\cos 	heta = \frac{	an \beta}{	an \alpha}$.Using the identity $	an^2 \frac{	heta}{

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

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

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

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