If x = sec theta - cos theta and y = sec^n th | vedclass

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(A) Given $x = \sec 	heta - \cos 	heta$ and $y = \sec^n 	heta - \cos^n 	heta$. We know that $(\sec^n 	heta + \cos^n 	heta)^2 = (\sec^n 	heta -

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$\frac{n^2(y^2 + 4)}{x^2 + 4}$

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(A) Given $x = \sec 	heta - \cos 	heta$ and $y = \sec^n 	heta - \cos^n 	heta$. We know that $(\sec^n 	heta + \cos^n 	heta)^2 = (\sec^n 	heta - \cos^n 	heta)^2 + 4(\sec^n 	heta \cdot \cos^n 	heta) = y^2 + 4$. Similarly,$(\sec 	heta + \cos 	heta)^2 = x^2 + 4$. Now,$\frac{dy}{d	heta} = n \sec^{n-1} 	heta \cdot \sec 	heta 	an 	heta - n \cos^{n-1} 	heta \cdot (-\sin 	heta) = n 	an 	heta (\sec^n 	heta + \cos^n 	heta)$. And $\frac{dx}{d	heta} = \sec 	heta 	an 	heta + \sin 	heta = 	an 	heta (\sec 	heta + \cos 	heta)$. Therefore,$\frac{dy}{dx} = \frac{n 	an 	heta (\sec^n 	heta + \cos^n 	heta)}{	an 	heta (\sec 	heta + \cos 	heta)} = n \frac{\sec^n 	heta + \cos^n 	heta}{\sec 	heta + \cos 	heta}$. Squaring both sides,$\left(\frac{dy}{dx}ight)^2 = n^2 \frac{(\sec^n 	heta + \cos^n 	heta)^2}{(\sec 	heta + \cos 	heta)^2} = \frac{n^2(y^2 + 4)}{x^2 + 4}$.

If $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$,then $\left(\frac{dy}{dx}\right)^2$ is equal to

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