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