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

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(A) Given $x = \sec 	heta - \cos 	heta$ and $y = \sec^n 	heta - \cos^n 	heta$. First,find the derivatives with respect to $	heta$: $\frac{dx}{d

Vedclass · Accepted Answer

$(x^2 + 4) \left( \frac{dy}{dx} \right)^2 = n^2(y^2 + 4)$

Vedclass · Answer

(A) Given $x = \sec 	heta - \cos 	heta$ and $y = \sec^n 	heta - \cos^n 	heta$. First,find the derivatives with respect to $	heta$: $\frac{dx}{d	heta} = \sec 	heta 	an 	heta + \sin 	heta = 	an 	heta (\sec 	heta + \cos 	heta)$. $\frac{dy}{d	heta} = n \sec^{n-1} 	heta (\sec 	heta 	an 	heta) + n \cos^{n-1} 	heta \sin 	heta = n 	an 	heta (\sec^n 	heta + \cos^n 	heta)$. Now,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{n 	an 	heta (\sec^n 	heta + \cos^n 	heta)}{	an 	heta (\sec 	heta + \cos 	heta)} = \frac{n(\sec^n 	heta + \cos^n 	heta)}{\sec 	heta + \cos 	heta}$. Squaring both sides: $\left( \frac{dy}{dx} ight)^2 = \frac{n^2(\sec^n 	heta + \cos^n 	heta)^2}{(\sec 	heta + \cos 	heta)^2}$. Using the identity $(a+b)^2 = (a-b)^2 + 4ab$: $(\sec^n 	heta + \cos^n 	heta)^2 = (\sec^n 	heta - \cos^n 	heta)^2 + 4 \sec^n 	heta \cos^n 	heta = y^2 + 4$. Similarly,$(\sec 	heta + \cos 	heta)^2 = (\sec 	heta - \cos 	heta)^2 + 4 \sec 	heta \cos 	heta = x^2 + 4$. Thus,$\left( \frac{dy}{dx} ight)^2 = \frac{n^2(y^2 + 4)}{x^2 + 4}$,which implies $(x^2 + 4) \left( \frac{dy}{dx} ight)^2 = n^2(y^2 + 4)$.

If $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$,then:

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