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

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(B) Given $x=\sec 	heta-\cos 	heta$ and $y=\sec ^n 	heta-\cos ^n 	heta$.First,differentiate $x$ with respect to $	heta$:$\frac{d x}{d 	heta} = \

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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 $y=\sec ^n 	heta-\cos ^n 	heta$.First,differentiate $x$ with respect to $	heta$:$\frac{d x}{d 	heta} = \sec 	heta 	an 	heta + \sin 	heta = \sec 	heta \cdot \frac{\sin 	heta}{\cos 	heta} + \sin 	heta = 	an 	heta(\sec 	heta + \cos 	heta)$.Next,differentiate $y$ with respect to $	heta$:$\frac{d y}{d 	heta} = n \sec ^{n-1} 	heta \cdot \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)$.Now,find $\frac{d y}{d x} = \frac{d y / d 	heta}{d x / d 	heta}$:$\frac{d y}{d x} = \frac{n 	an 	heta(\sec ^n 	heta + \cos ^n 	heta)}{	an 	heta(\sec 	heta + \cos 	heta)} = \frac{n(\sec ^n 	heta + \cos ^n 	heta)}{\sec 	heta + \cos 	heta}$.Squaring both sides:$\left(\frac{d y}{d x}ight)^2 = \frac{n^2(\sec ^n 	heta + \cos ^n 	heta)^2}{(\sec 	heta + \cos 	heta)^2} = \frac{n^2(\sec ^{2n} 	heta + \cos ^{2n} 	heta + 2)}{(\sec ^2 	heta + \cos ^2 	heta + 2)}$.Since $(\sec ^n 	heta - \cos ^n 	heta)^2 = \sec ^{2n} 	heta + \cos ^{2n} 	heta - 2$,we have $\sec ^{2n} 	heta + \cos ^{2n} 	heta = y^2 + 2$.Substituting this:$\left(\frac{d y}{d x}ight)^2 = \frac{n^2(y^2 + 2 + 2)}{x^2 + 2 + 2} = \frac{n^2(y^2 + 4)}{x^2 + 4}$.Therefore,$\frac{d y}{d x} = n \sqrt{\frac{y^2+4}{x^2+4}}$.

If $x=\sec \theta-\cos \theta$ and $y=\sec ^n \theta-\cos ^n \theta$,then $\frac{d y}{d x}=$

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