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

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

$\begin{array}{c}\frac{dx}{d\theta }=a\left[-\sin \theta +\frac{1}{\tan \theta /2}{\sec }^2\frac{\theta }{2}\times \frac{1}{2}\right]\\ =a\left[-\sin \theta +\frac{1}{2\sin (\theta /2)\cos (\theta /2)}\right]\\ =a\left[-\sin \theta +\frac{1}{\sin \theta }\right]\end{array}$

$\frac{dy}{d\theta }=a\cos \theta$

$\therefore \frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{a\cos \theta }{\frac{a{\cos }^2\theta }{\sin \theta }}=\tan \theta$