If x=operatorname{cosec} theta-sin theta, y=o | vedclass

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(B) Given, $x=\operatorname{cosec} 	heta-\sin 	heta$. Then, $\frac{d x}{d 	heta}=-\operatorname{cosec} 	heta \cot 	heta-\cos 	heta=-\cot 	heta(

Vedclass · Accepted Answer

$6$

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(B) Given, $x=\operatorname{cosec} 	heta-\sin 	heta$. Then, $\frac{d x}{d 	heta}=-\operatorname{cosec} 	heta \cot 	heta-\cos 	heta=-\cot 	heta(\operatorname{cosec} 	heta+\sin 	heta)$. Since $x^2+4=(\operatorname{cosec} 	heta-\sin 	heta)^2+4=\operatorname{cosec}^2 	heta+\sin^2 	heta-2+\dots+4=(\operatorname{cosec} 	heta+\sin 	heta)^2$, we have $\sqrt{x^2+4}=\operatorname{cosec} 	heta+\sin 	heta$. Thus, $\frac{d x}{d 	heta}=-\cot 	heta \sqrt{x^2+4}$. Similarly, $y=\operatorname{cosec}^{2022} 	heta-\sin ^{2022} 	heta$. Then, $\frac{d y}{d 	heta}=-2022 \operatorname{cosec}^{2021} 	heta \operatorname{cosec} 	heta \cot 	heta-2022 \sin ^{2021} 	heta \cos 	heta = -2022 \cot 	heta(\operatorname{cosec}^{2022} 	heta+\sin^{2022} 	heta)$. Since $y^2+4=(\operatorname{cosec}^{2022} 	heta-\sin^{2022} 	heta)^2+4=(\operatorname{cosec}^{2022} 	heta+\sin^{2022} 	heta)^2$, we have $\sqrt{y^2+4}=\operatorname{cosec}^{2022} 	heta+\sin^{2022} 	heta$. Thus, $\frac{d y}{d 	heta}=-2022 \cot 	heta \sqrt{y^2+4}$. Dividing the two derivatives, $\frac{d y}{d x} = \frac{d y / d 	heta}{d x / d 	heta} = \frac{-2022 \cot 	heta \sqrt{y^2+4}}{-\cot 	heta \sqrt{x^2+4}} = 2022 \frac{\sqrt{y^2+4}}{\sqrt{x^2+4}}$. Squaring both sides, $\left(\frac{d y}{d x}ight)^2 = (2022)^2 \frac{y^2+4}{x^2+4}$. Comparing this with $\frac{k(y^2+4)}{g(x)}$, we get $k=(2022)^2$ and $g(x)=x^2+4$. Finally, $10+k-g(2022) = 10+(2022)^2-(2022^2+4) = 10-4 = 6$.

List-$I$	List-$II$
$(i)$ $x = a(\theta - \sin \theta), y = a(1 - \cos \theta)$	$(A)$ $4\sqrt{3}$
(ii) $x = 3\cos \theta - 2\cos^3 \theta, y = 3\sin \theta - 2\sin^3 \theta$	$(B)$ $-\frac{1}{3\sqrt{3}}$
(iii) $x = 3\cos \theta - \cos^3 \theta, y = 3\sin \theta - \sin^3 \theta$	$(C)$ $\sqrt{3}$
(iv) $x = a \log \sin \theta, y = a \tan \theta$	$(D)$ $\frac{1}{\sqrt{3}}$
	$(E)$ $\frac{1}{3\sqrt{3}}$

If $x=\operatorname{cosec} \theta-\sin \theta$, $y=\operatorname{cosec}^{2022} \theta-\sin ^{2022} \theta$ and $\left(\frac{d y}{d x}\right)^2=\frac{k\left(y^2+4\right)}{g(x)}$ where $k \in R$, then $10+k-g(2022)=$

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