If a point (x, y) equiv (tan theta + sin thet | vedclass

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(C) Given: $x = 	an 	heta + \sin 	heta$ and $y = 	an 	heta - \sin 	heta$.Step $1$: Calculate $x + y$ and $x - y$.$x + y = 2 	an 	heta$$x - y =

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$({x^2} - {y^2})^2 = 16xy$

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(C) Given: $x = 	an 	heta + \sin 	heta$ and $y = 	an 	heta - \sin 	heta$.Step $1$: Calculate $x + y$ and $x - y$.$x + y = 2 	an 	heta$$x - y = 2 \sin 	heta$Step $2$: Express $	an 	heta$ and $\sin 	heta$ in terms of $x$ and $y$.$	an 	heta = \frac{x + y}{2}$$\sin 	heta = \frac{x - y}{2}$Step $3$: Use the identity $\frac{1}{\sin^2 	heta} - \frac{1}{	an^2 	heta} = \frac{1 - \sin^2 	heta}{\sin^2 	heta} = \frac{\cos^2 	heta}{\sin^2 	heta} = \cot^2 	heta$ is not directly helpful. Instead,use $\frac{1}{\sin^2 	heta} - \frac{1}{	an^2 	heta} = \frac{1}{\sin^2 	heta} - \frac{\cos^2 	heta}{\sin^2 	heta} = \frac{1 - \cos^2 	heta}{\sin^2 	heta} = 1$.Step $4$: Substitute the values:$\frac{4}{(x - y)^2} - \frac{4}{(x + y)^2} = 1$$4 \left[ \frac{(x + y)^2 - (x - y)^2}{(x - y)^2 (x + y)^2} ight] = 1$$4 \left[ \frac{4xy}{(x^2 - y^2)^2} ight] = 1$$16xy = (x^2 - y^2)^2$.Thus,the locus is $(x^2 - y^2)^2 = 16xy$.

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

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