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

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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 $x = \frac{1}{\cos 	heta} - \cos 	heta = \frac{1 - \cos^2 	heta}{\cos 	heta} = \frac{\sin^2 	heta}{\cos 	heta}$. Also,$x^2 + 4 = (\sec 	heta - \cos 	heta)^2 + 4 = \sec^2 	heta - 2 + \cos^2 	heta + 4 = \sec^2 	heta + 2 + \cos^2 	heta = (\sec 	heta + \cos 	heta)^2$. Thus,$\sqrt{x^2 + 4} = \sec 	heta + \cos 	heta$. Similarly,$y^2 + 4 = (\sec^n 	heta - \cos^n 	heta)^2 + 4 = \sec^{2n} 	heta - 2 + \cos^{2n} 	heta + 4 = \sec^{2n} 	heta + 2 + \cos^{2n} 	heta = (\sec^n 	heta + \cos^n 	heta)^2$. Thus,$\sqrt{y^2 + 4} = \sec^n 	heta + \cos^n 	heta$. Now,$\frac{dx}{d	heta} = \sec 	heta 	an 	heta + \sin 	heta = 	an 	heta (\sec 	heta + \cos 	heta)$. And $\frac{dy}{d	heta} = n \sec^{n-1} 	heta (\sec 	heta 	an 	heta) - n \cos^{n-1} 	heta (-\sin 	heta) = n \sec^n 	heta 	an 	heta + n \cos^{n-1} 	heta \sin 	heta = n 	an 	heta (\sec^n 	heta + \cos^n 	heta)$. Therefore,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{n 	an 	heta (\sec^n 	heta + \cos^n 	heta)}{	an 	heta (\sec 	heta + \cos 	heta)} = n \frac{\sec^n 	heta + \cos^n 	heta}{\sec 	heta + \cos 	heta} = n \frac{\sqrt{y^2 + 4}}{\sqrt{x^2 + 4}} = n \sqrt{\frac{y^2 + 4}{x^2 + 4}}$.

If $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$,then find $\frac{dy}{dx}$.

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