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 equations,we get:$	an 	heta = \frac{x

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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 equations,we get:$	an 	heta = \frac{x + y}{2}$ and $\sin 	heta = \frac{x - y}{2}$.Now,consider the product $xy$:$xy = (	an 	heta + \sin 	heta)(	an 	heta - \sin 	heta) = 	an^2 	heta - \sin^2 	heta$.Using $	an^2 	heta = \frac{\sin^2 	heta}{\cos^2 	heta}$,we have:$xy = \sin^2 	heta \left(\frac{1}{\cos^2 	heta} - 1ight) = \sin^2 	heta \left(\frac{1 - \cos^2 	heta}{\cos^2 	heta}ight) = \sin^2 	heta \cdot 	an^2 	heta$.Also,$x^2 - y^2 = (	an 	heta + \sin 	heta)^2 - (	an 	heta - \sin 	heta)^2 = 4 	an 	heta \sin 	heta$.Squaring both sides:$(x^2 - y^2)^2 = 16 	an^2 	heta \sin^2 	heta$.Since $xy = 	an^2 	heta \sin^2 	heta$,we get:$(x^2 - y^2)^2 = 16xy$.

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

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