If tan theta = frac{x sin phi}{1 - x cos phi} | vedclass

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(B) Given $	an 	heta = \frac{x \sin \phi}{1 - x \cos \phi}$.Rearranging,we get $x \sin \phi = 	an 	heta - x \cos \phi 	an 	heta$.$x(\sin \phi +

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$\frac{\sin 	heta}{\sin \phi}$

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(B) Given $	an 	heta = \frac{x \sin \phi}{1 - x \cos \phi}$.Rearranging,we get $x \sin \phi = 	an 	heta - x \cos \phi 	an 	heta$.$x(\sin \phi + \cos \phi 	an 	heta) = 	an 	heta$.$x \left( \sin \phi + \cos \phi \frac{\sin 	heta}{\cos 	heta} ight) = \frac{\sin 	heta}{\cos 	heta}$.$x \left( \frac{\sin \phi \cos 	heta + \cos \phi \sin 	heta}{\cos 	heta} ight) = \frac{\sin 	heta}{\cos 	heta}$.$x \sin(	heta + \phi) = \sin 	heta \Rightarrow x = \frac{\sin 	heta}{\sin(	heta + \phi)}$.Similarly,for $	an \phi = \frac{y \sin 	heta}{1 - y \cos 	heta}$,we get $y = \frac{\sin \phi}{\sin(	heta + \phi)}$.Therefore,$\frac{x}{y} = \frac{\sin 	heta / \sin(	heta + \phi)}{\sin \phi / \sin(	heta + \phi)} = \frac{\sin 	heta}{\sin \phi}$.

If $\tan \theta = \frac{x \sin \phi}{1 - x \cos \phi}$ and $\tan \phi = \frac{y \sin \theta}{1 - y \cos \theta}$,then $\frac{x}{y} = $

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