If $a=\tan 27\theta -\tan \theta$ and $b=\frac{\sin \theta }{\cos 3\theta }+\frac{\sin 3\theta }{\cos 9\theta }+\frac{\sin 9\theta }{\cos 27\theta },$ then

We have,

      $a=\tan 27\theta -\tan \theta$

$\begin{array}{l}= a=\tan 27\theta -\tan 9\theta +\tan 9\theta -\tan 3\theta +\tan 3\theta -\tan \theta \\ \Rightarrow a=\frac{\sin 18\theta }{\cos 27\theta \cos 9\theta }+\frac{\sin 6\theta }{\cos 9\theta \cos 3\theta }+\frac{\sin 2\theta }{\cos 3\theta \cos \theta }\end{array}$

$\Rightarrow a=2\left(\frac{\sin 9\theta }{\cos 27\theta }+\frac{\sin 3\theta }{\cos 9\theta }+\frac{\sin \theta }{\cos 3\theta }\right)\Rightarrow a=2b$