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

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(B) Given $x = \frac{2t}{1 + t^2}$ and $y = \frac{1 - t^2}{1 + t^2}$.Substitute $t = 	an 	heta$.Then $x = \frac{2 	an 	heta}{1 + 	an^2 	heta} =

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$\frac{2t}{t^2 - 1}$

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(B) Given $x = \frac{2t}{1 + t^2}$ and $y = \frac{1 - t^2}{1 + t^2}$.Substitute $t = 	an 	heta$.Then $x = \frac{2 	an 	heta}{1 + 	an^2 	heta} = \sin 2	heta$.And $y = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta} = \cos 2	heta$.Now,find the derivatives with respect to $	heta$:$\frac{dx}{d	heta} = 2 \cos 2	heta$.$\frac{dy}{d	heta} = -2 \sin 2	heta$.Using the chain rule,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{-2 \sin 2	heta}{2 \cos 2	heta} = -	an 2	heta$.Since $	an 2	heta = \frac{2 	an 	heta}{1 - 	an^2 	heta} = \frac{2t}{1 - t^2}$,$\frac{dy}{dx} = -\left( \frac{2t}{1 - t^2} ight) = \frac{2t}{t^2 - 1}$.

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

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