If x = operatorname{cosec} theta - sin theta | vedclass

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(B) Given: $x = \operatorname{cosec} 	heta - \sin 	heta = \frac{1}{\sin 	heta} - \sin 	heta = \frac{1 - \sin^{2} 	heta}{\sin 	heta} = \frac{\cos

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$x^{2} y^{2} (x^{2} + y^{2} + 3) = 1$

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(B) Given: $x = \operatorname{cosec} 	heta - \sin 	heta = \frac{1}{\sin 	heta} - \sin 	heta = \frac{1 - \sin^{2} 	heta}{\sin 	heta} = \frac{\cos^{2} 	heta}{\sin 	heta}$.Given: $y = \sec 	heta - \cos 	heta = \frac{1}{\cos 	heta} - \cos 	heta = \frac{1 - \cos^{2} 	heta}{\cos 	heta} = \frac{\sin^{2} 	heta}{\cos 	heta}$.Now,consider the expression $x^{2} y^{2} (x^{2} + y^{2} + 3)$:$x^{2} y^{2} = \left( \frac{\cos^{4} 	heta}{\sin^{2} 	heta} ight) \left( \frac{\sin^{4} 	heta}{\cos^{2} 	heta} ight) = \cos^{2} 	heta \sin^{2} 	heta$.$x^{2} + y^{2} + 3 = \frac{\cos^{4} 	heta}{\sin^{2} 	heta} + \frac{\sin^{4} 	heta}{\cos^{2} 	heta} + 3 = \frac{\cos^{6} 	heta + \sin^{6} 	heta + 3 \sin^{2} 	heta \cos^{2} 	heta}{\sin^{2} 	heta \cos^{2} 	heta}$.Using the identity $a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})$,let $a = \cos^{2} 	heta$ and $b = \sin^{2} 	heta$:$\cos^{6} 	heta + \sin^{6} 	heta = (\cos^{2} 	heta + \sin^{2} 	heta)(\cos^{4} 	heta - \cos^{2} 	heta \sin^{2} 	heta + \sin^{4} 	heta) = 1 \cdot (\cos^{4} 	heta - \cos^{2} 	heta \sin^{2} 	heta + \sin^{4} 	heta)$.Substituting this back: $x^{2} + y^{2} + 3 = \frac{(\cos^{4} 	heta - \cos^{2} 	heta \sin^{2} 	heta + \sin^{4} 	heta) + 3 \sin^{2} 	heta \cos^{2} 	heta}{\sin^{2} 	heta \cos^{2} 	heta} = \frac{\cos^{4} 	heta + 2 \sin^{2} 	heta \cos^{2} 	heta + \sin^{4} 	heta}{\sin^{2} 	heta \cos^{2} 	heta} = \frac{(\cos^{2} 	heta + \sin^{2} 	heta)^{2}}{\sin^{2} 	heta \cos^{2} 	heta} = \frac{1}{\sin^{2} 	heta \cos^{2} 	heta}$.Therefore,$x^{2} y^{2} (x^{2} + y^{2} + 3) = (\cos^{2} 	heta \sin^{2} 	heta) \cdot \left( \frac{1}{\sin^{2} 	heta \cos^{2} 	heta} ight) = 1$.

If $x = \operatorname{cosec} \theta - \sin \theta$ and $y = \sec \theta - \cos \theta$,then the relation between $x$ and $y$ is

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