Solve this

We have, $x=2 \cos \theta-\cos 2 \theta$

$\Rightarrow \frac{d x}{d \theta}=2(-\sin \theta)-(-\sin 2 \theta) \frac{d}{d \theta}(2 \theta)$

$\Rightarrow \frac{d x}{d \theta}=-2 \sin \theta+2 \sin 2 \theta$

$\Rightarrow \frac{d x}{d \theta}=2(\sin 2 \theta-\sin \theta)$                      .....(1)

and,

$y=2 \sin \theta-\sin 2 \theta$

$\Rightarrow \frac{d y}{d \theta}=2 \cos \theta-\cos 2 \theta \frac{d}{d \theta}(2 \theta)$

$\Rightarrow \frac{d y}{d \theta}=2 \cos \theta-\cos 2 \theta(2)$

$\Rightarrow \frac{d y}{d \theta}=2 \cos \theta-2 \cos 2 \theta$

$\Rightarrow \frac{d y}{d \theta}=2(\cos \theta-\cos 2 \theta)$                  ......(2)

Dividing equation (ii) by equation (i),

$\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{2(\cos \theta-\cos 2 \theta)}{2(\sin 2 \theta-\sin \theta)}$

$\Rightarrow \frac{d y}{d x}=\frac{\cos \theta-\cos 2 \theta}{\sin 2 \theta-\sin \theta}$

$\Rightarrow \frac{d y}{d x}=\frac{-2 \sin \left(\frac{\theta+2 \theta}{2}\right) \sin \left(\frac{\theta-2 \theta}{2}\right)}{2 \cos \left(\frac{2 \theta+\theta}{2}\right) \sin \left(\frac{2 \theta-\theta}{2}\right)}$                              $\left[\begin{array}{c}\because \sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right) \\ \text { and } \cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\end{array}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{-\sin \left(\frac{3 \theta}{2}\right) \sin \left(\frac{-\theta}{2}\right)}{\cos \left(\frac{3 \theta}{2}\right) \sin \left(\frac{\theta}{2}\right)}$

$\Rightarrow \frac{d y}{d x}=\frac{-\sin \left(\frac{3 \theta}{2}\right)\left(-\sin \frac{\theta}{2}\right)}{\cos \left(\frac{3 \theta}{2}\right) \sin \left(\frac{\theta}{2}\right)}$

$\Rightarrow \frac{d y}{d x}=\frac{\sin \left(\frac{3 \theta}{2}\right)}{\cos \left(\frac{3 \theta}{2}\right)}$

$\Rightarrow \frac{d y}{d x}=\tan \left(\frac{3 \theta}{2}\right)$