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 $x$ with respect to $	heta$: $\frac{dx}{d	heta} =

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(D) Given $x=\sec 	heta-\cos 	heta$ and $y=\sec ^{10} 	heta-\cos ^{10} 	heta$. Differentiating $x$ with respect to $	heta$: $\frac{dx}{d	heta} = \sec 	heta 	an 	heta + \sin 	heta = 	an 	heta (\sec 	heta + \cos 	heta)$. Differentiating $y$ with respect to $	heta$: $\frac{dy}{d	heta} = 10 \sec^9 	heta (\sec 	heta 	an 	heta) - 10 \cos^9 	heta (-\sin 	heta) = 10 	an 	heta (\sec^{10} 	heta + \cos^{10} 	heta)$. Thus,$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = 10 \frac{\sec^{10} 	heta + \cos^{10} 	heta}{\sec 	heta + \cos 	heta}$. Squaring both sides: $(\frac{dy}{dx})^2 = 100 \frac{(\sec^{10} 	heta + \cos^{10} 	heta)^2}{(\sec 	heta + \cos 	heta)^2}$. Using the identity $(a+b)^2 = (a-b)^2 + 4ab$: $(\sec^{10} 	heta + \cos^{10} 	heta)^2 = (\sec^{10} 	heta - \cos^{10} 	heta)^2 + 4(\sec^{10} 	heta \cos^{10} 	heta) = y^2 + 4$. Similarly,$(\sec 	heta + \cos 	heta)^2 = (\sec 	heta - \cos 	heta)^2 + 4 = x^2 + 4$. Substituting these into the expression for $(\frac{dy}{dx})^2$: $(\frac{dy}{dx})^2 = 100 \frac{y^2+4}{x^2+4}$. Therefore,$(x^2+4)(\frac{dy}{dx})^2 = 100(y^2+4)$. Comparing this with the given equation $(x^2+4)(\frac{dy}{dx})^2 = k(y^2+4)$,we get $k=100$.

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 $k=$

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