If x = tan^{-1} left{ frac{sqrt{1+t^2}-1}{t} | vedclass

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(C) Let $t = 	an 	heta$. Then $	heta = 	an^{-1} t$. For $x$: $x = 	an^{-1} \left\{ \frac{\sqrt{1+	an^2 	heta}-1}{	an 	heta} ight\} = 	an^{

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$4$

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(C) Let $t = 	an 	heta$. Then $	heta = 	an^{-1} t$. For $x$: $x = 	an^{-1} \left\{ \frac{\sqrt{1+	an^2 	heta}-1}{	an 	heta} ight\} = 	an^{-1} \left\{ \frac{\sec 	heta - 1}{	an 	heta} ight\} = 	an^{-1} \left\{ \frac{1-\cos 	heta}{\sin 	heta} ight\} = 	an^{-1} \left\{ 	an \frac{	heta}{2} ight\} = \frac{	heta}{2} = \frac{1}{2} 	an^{-1} t$. Thus,$\frac{dx}{dt} = \frac{1}{2(1+t^2)}$. For $y$: $y = \cos^{-1} \left\{ \frac{1-	an^2 	heta}{1+	an^2 	heta} ight\} = \cos^{-1} (\cos 2	heta) = 2	heta = 2 	an^{-1} t$. Thus,$\frac{dy}{dt} = \frac{2}{1+t^2}$. Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2/(1+t^2)}{1/(2(1+t^2))} = 2 	imes 2 = 4$.

If $x = \tan^{-1} \left\{ \frac{\sqrt{1+t^2}-1}{t} \right\}$ and $y = \cos^{-1} \left\{ \frac{1-t^2}{1+t^2} \right\}$,then $\frac{dy}{dx}$ is equal to

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