If x = frac{1 - t^2}{1 + t^2} and y = frac{2t | vedclass

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(C) Given $x = \frac{1 - t^2}{1 + t^2}$ and $y = \frac{2t}{1 + t^2}$.Substitute $t = 	an 	heta$ in both equations:$x = \frac{1 - 	an^2 	heta}{1 +

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$-\frac{x}{y}$

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(C) Given $x = \frac{1 - t^2}{1 + t^2}$ and $y = \frac{2t}{1 + t^2}$.Substitute $t = 	an 	heta$ in both equations:$x = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta} = \cos 2	heta$$y = \frac{2 	an 	heta}{1 + 	an^2 	heta} = \sin 2	heta$Differentiating both with respect to $	heta$:$\frac{dx}{d	heta} = -2 \sin 2	heta$$\frac{dy}{d	heta} = 2 \cos 2	heta$Using the chain rule,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{2 \cos 2	heta}{-2 \sin 2	heta} = -\cot 2	heta$.Since $x = \cos 2	heta$ and $y = \sin 2	heta$,we have $\frac{dy}{dx} = -\frac{\cos 2	heta}{\sin 2	heta} = -\frac{x}{y}$.

If $x = \frac{1 - t^2}{1 + t^2}$ and $y = \frac{2t}{1 + t^2}$,then $\frac{dy}{dx} = $

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