If a point (x, y) = (tan theta + sin theta, t | vedclass

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(D) Given that,$x = 	an 	heta + \sin 	heta$ and $y = 	an 	heta - \sin 	heta$.Adding and subtracting the two equations,we get:$	an 	heta = \fra

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$\left(x^2 - y^2\right)^2 = 16xy$

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(D) Given that,$x = 	an 	heta + \sin 	heta$ and $y = 	an 	heta - \sin 	heta$.Adding and subtracting the two equations,we get:$	an 	heta = \frac{x + y}{2}$ and $\sin 	heta = \frac{x - y}{2}$.Now,consider the expression $x^2 - y^2$:$x^2 - y^2 = (x + y)(x - y) = (2 	an 	heta)(2 \sin 	heta) = 4 	an 	heta \sin 	heta$.We know that $	an 	heta \sin 	heta = \frac{\sin^2 	heta}{\cos 	heta} = \frac{1 - \cos^2 	heta}{\cos 	heta} = \sec 	heta - \cos 	heta$.However,a more direct approach is:$(x^2 - y^2)^2 = (4 	an 	heta \sin 	heta)^2 = 16 	an^2 	heta \sin^2 	heta$.Since $\sin^2 	heta = 	an^2 	heta \cos^2 	heta$,we have $\sin^2 	heta = 	an^2 	heta (1 - \sin^2 	heta)$,which implies $\sin^2 	heta (1 + 	an^2 	heta) = 	an^2 	heta$,so $\sin^2 	heta = \frac{	an^2 	heta}{\sec^2 	heta}$.Alternatively,note that $xy = (	an 	heta + \sin 	heta)(	an 	heta - \sin 	heta) = 	an^2 	heta - \sin^2 	heta = 	an^2 	heta (1 - \cos^2 	heta) = 	an^2 	heta \sin^2 	heta$.Thus,$(x^2 - y^2)^2 = 16 	an^2 	heta \sin^2 	heta = 16xy$.

If a point $(x, y) = (\tan \theta + \sin \theta, \tan \theta - \sin \theta)$,then the locus of $(x, y)$ is

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