If $x=a\left\{\cos \theta+\log \tan \frac{\theta}{2}\right\}$ and $y=a \sin \theta$, then $\frac{d y}{d x}$ is equal to

$\tan \theta$

We have,

$x=a\left\{\cos \theta+\log \tan \frac{\theta}{2}\right\}$ and $y=a \sin \theta$

$\Rightarrow \frac{d x}{d \theta}=a\left\{-\sin \theta+\frac{\sec ^2 \frac{\theta}{2}}{2 \tan \frac{\theta}{2}}\right\}$ and, $\frac{d y}{d \theta}=a \cos \theta$

$\Rightarrow \frac{d x}{d \theta}=a\left\{-\sin \theta+\frac{1}{\sin \theta}\right\}$ and, $\frac{d y}{d \theta}=a \cos \theta$

$\Rightarrow \frac{d x}{d \theta}=a \frac{\cos ^2 \theta}{\sin \theta}$ and, $\frac{d y}{d \theta}=a \cos \theta$

∴  $\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{a \cos \theta}{a \frac{\cos ^2 \theta}{\sin \theta}}=\tan \theta$