Find $\frac{d y}{d x}$, when
$x=a(\cos \theta+\theta \sin \theta)$ and $y=a(\sin \theta-\cos \theta)$
the given equation are $x=a(\cos \theta+\theta \sin \theta)$
Then $\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}\left[\frac{\mathrm{d}}{\mathrm{d} \theta} \cos \theta+\frac{\mathrm{d}}{\mathrm{d} \theta}(\theta \sin \theta)\right]$
$=a\left[-\sin \theta+\frac{\theta d}{d \theta}(\sin \theta)+\sin \theta \frac{d}{d \theta}(\theta)\right]$
$=a[-\sin \theta+\theta \cos \theta+\sin \theta]=a \theta \cos \theta$
And $y=a(\sin \theta-\cos \theta)$ so
$\frac{d y}{d \theta}=a\left[\frac{d}{d \theta}(\sin \theta)-\frac{d}{d \theta}(\theta \cos \theta)\right]$
$=a\left[\cos \theta-\left\{\frac{\theta d}{d \theta}(\cos \theta)+\cos \theta \frac{d}{d \theta}(\theta)\right\}\right]$
$=a[\cos \theta+\theta \sin \theta-\cos \theta]$
$=a \theta \sin \theta$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{d} \theta}}{\frac{\mathrm{dx}}{\mathrm{d} \theta}}=\frac{\mathrm{a} \theta \sin \theta}{\mathrm{a} \theta \cos \theta}=\tan \theta$