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

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(A) Given $x = \sec 	heta - \cos 	heta$ and $y = \sec^n 	heta - \cos^n 	heta$. Let $u = \sec 	heta$ and $v = \cos 	heta = \frac{1}{u}$. Then $x

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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 $y = \sec^n 	heta - \cos^n 	heta$. Let $u = \sec 	heta$ and $v = \cos 	heta = \frac{1}{u}$. Then $x = u - \frac{1}{u}$ and $y = u^n - \frac{1}{u^n}$. We know that $x^2 = (u - \frac{1}{u})^2 = u^2 + \frac{1}{u^2} - 2$,so $x^2 + 4 = u^2 + \frac{1}{u^2} + 2 = (u + \frac{1}{u})^2$. Thus,$u + \frac{1}{u} = \sqrt{x^2 + 4}$. Similarly,$y^2 + 4 = (u^n - \frac{1}{u^n})^2 + 4 = u^{2n} + \frac{1}{u^{2n}} - 2 + 4 = u^{2n} + \frac{1}{u^{2n}} + 2 = (u^n + \frac{1}{u^n})^2$. Thus,$u^n + \frac{1}{u^n} = \sqrt{y^2 + 4}$. Using the identity $(u^n - \frac{1}{u^n})^2 = (u^n + \frac{1}{u^n})^2 - 4$,we have $y^2 = (y^2 + 4) - 4$. Now,differentiate $x$ and $y$ with respect to $	heta$: $\frac{dx}{d	heta} = \sec 	heta 	an 	heta + \sin 	heta = \sec 	heta 	an 	heta + \sin 	heta = 	an 	heta (\sec 	heta + \cos 	heta)$. $\frac{dy}{d	heta} = n \sec^{n-1} 	heta (\sec 	heta 	an 	heta) + n \cos^{n-1} 	heta \sin 	heta = n 	an 	heta (\sec^n 	heta + \cos^n 	heta)$. Then $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{n(\sec^n 	heta + \cos^n 	heta)}{\sec 	heta + \cos 	heta} = \frac{n \sqrt{y^2 + 4}}{\sqrt{x^2 + 4}}$. Squaring both sides,we get ${\left( {\frac{{dy}}{{dx}}} ight)^2} = \frac{{n^2(y^2 + 4)}}{{x^2 + 4}}$.

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

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