If $x=a(\theta+\sin \theta), y=a(1+\cos \theta)$, find $\frac{d y}{d x}$
$\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}(1+\cos \theta)$ and $\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}(-\sin \theta)$
Using Chain Rule of Differentiation,
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{d} \theta} \cdot \frac{\mathrm{d} \theta}{\mathrm{dx}}$
$=a(-\sin \theta) \cdot \frac{1}{a(1+\cos \theta)}$
$=-\frac{\sin \theta}{1+\cos \theta}$
$=-\frac{\sin \theta}{1+\cos \theta} \cdot \frac{1-\cos \theta}{1-\cos \theta}$
$=-\frac{\sin \theta(1-\cos \theta)}{1-\cos ^{2} \theta}$
$=-\frac{\sin \theta(1-\cos \theta)}{\sin ^{2} \theta}$
$=-\frac{1-\cos \theta}{\sin \theta}$
$=\cot \theta-\operatorname{cosec} \theta($ Ans $)$
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