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

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(B) Given $x = \sec 	heta - \cos 	heta$ and $y = \sec^n 	heta - \cos^n 	heta$. We know that $\sec 	heta = \frac{1}{\cos 	heta}$,so $x = \frac{1}

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$n^2(y^2 + 4)$

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(B) Given $x = \sec 	heta - \cos 	heta$ and $y = \sec^n 	heta - \cos^n 	heta$. We know that $\sec 	heta = \frac{1}{\cos 	heta}$,so $x = \frac{1}{\cos 	heta} - \cos 	heta = \frac{\sin^2 	heta}{\cos 	heta}$. Then $\frac{dx}{d	heta} = \sec 	heta 	an 	heta + \sin 	heta = \frac{\sin 	heta}{\cos^2 	heta} + \sin 	heta = \sin 	heta \left(\frac{1 + \cos^2 	heta}{\cos^2 	heta}ight)$. Similarly,$\frac{dy}{d	heta} = n \sec^n 	heta 	an 	heta + n \cos^{n-1} 	heta \sin 	heta = n \sin 	heta \left(\frac{1 + \cos^{2n} 	heta}{\cos^{n+1} 	heta}ight)$. Now,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{n \sin 	heta (1 + \cos^{2n} 	heta) / \cos^{n+1} 	heta}{\sin 	heta (1 + \cos^2 	heta) / \cos^2 	heta} = \frac{n (1 + \cos^{2n} 	heta)}{\cos^{n-1} 	heta (1 + \cos^2 	heta)}$. Also,$x^2 + 4 = (\sec 	heta - \cos 	heta)^2 + 4 = \sec^2 	heta - 2 + \cos^2 	heta + 4 = \sec^2 	heta + 2 + \cos^2 	heta = (\sec 	heta + \cos 	heta)^2 = \left(\frac{1 + \cos^2 	heta}{\cos 	heta}ight)^2$. Thus,$(x^2 + 4) \left(\frac{dy}{dx}ight)^2 = \left(\frac{1 + \cos^2 	heta}{\cos 	heta}ight)^2 \cdot \frac{n^2 (1 + \cos^{2n} 	heta)^2}{\cos^{2n-2} 	heta (1 + \cos^2 	heta)^2} = \frac{n^2 (1 + \cos^{2n} 	heta)^2}{\cos^{2n} 	heta} = n^2 \left(\frac{1}{\cos^n 	heta} - \cos^n 	hetaight)^2 + 4n^2 = n^2 (y^2 + 4)$.

If $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$,then $(x^2 + 4) \left(\frac{dy}{dx}\right)^2 =$

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