If x=sec theta-cos theta,y=sec ^{10} theta-co | vedclass

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(D) Given $x=\sec 	heta-\cos 	heta$ and $y=\sec ^{10} 	heta-\cos ^{10} 	heta$.Differentiating with respect to $	heta$:$\frac{dx}{d	heta} = \sec

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(D) Given $x=\sec 	heta-\cos 	heta$ and $y=\sec ^{10} 	heta-\cos ^{10} 	heta$.Differentiating with respect to $	heta$:$\frac{dx}{d	heta} = \sec 	heta 	an 	heta + \sin 	heta = 	an 	heta (\sec 	heta + \cos 	heta)$$\frac{dy}{d	heta} = 10 \sec^9 	heta (\sec 	heta 	an 	heta) - 10 \cos^9 	heta (-\sin 	heta) = 10 \sec^{10} 	heta 	an 	heta + 10 \cos^9 	heta \sin 	heta$$= 10 	an 	heta (\sec^{10} 	heta + \cos^{10} 	heta)$Now,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{10 	an 	heta (\sec^{10} 	heta + \cos^{10} 	heta)}{	an 	heta (\sec 	heta + \cos 	heta)} = \frac{10 (\sec^{10} 	heta + \cos^{10} 	heta)}{\sec 	heta + \cos 	heta}$Squaring both sides:$(\frac{dy}{dx})^2 = \frac{100 (\sec^{10} 	heta + \cos^{10} 	heta)^2}{(\sec 	heta + \cos 	heta)^2}$Using the identity $(a+b)^2 = (a-b)^2 + 4ab$:$(\frac{dy}{dx})^2 = \frac{100 [(\sec^{10} 	heta - \cos^{10} 	heta)^2 + 4 \sec^{10} 	heta \cos^{10} 	heta]}{(\sec 	heta - \cos 	heta)^2 + 4 \sec 	heta \cos 	heta}$Since $\sec 	heta \cos 	heta = 1$,we have:$(\frac{dy}{dx})^2 = \frac{100 (y^2 + 4)}{x^2 + 4}$Thus,$(x^2+4)(\frac{dy}{dx})^2 = 100(y^2+4)$.Comparing with the given equation,$k = 100$.

If $x=\sec \theta-\cos \theta$,$y=\sec ^{10} \theta-\cos ^{10} \theta$ and $(x^2+4)(\frac{dy}{dx})^2=k(y^2+4)$,then the value of $k$ is

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