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Question:

Find $\frac{d y}{d x}$, when

$x=a(\theta+\sin \theta)$ and $y=a(1-\cos \theta)$

Solution:

We have, $x=a(\theta+\sin \theta)$ and $y=a(1-\cos \theta)$

$\Rightarrow \frac{d x}{d \theta}=a(1+\cos \theta)$ and $\frac{d y}{d \theta}=a \sin \theta$

$\therefore \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{a \sin \theta}{a(1+\cos \theta)}=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \theta}=\tan \frac{\theta}{2}$

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