If $\frac{\sin \theta }{\cos 3\theta }+\frac{\sin 3\theta }{\cos 9\theta }+\frac{\sin 9\theta }{\cos 27\theta }+\frac{\sin 27\theta }{\cos 81\theta }=A\left(\tan B\theta -\tan C\theta \right)$  then $\left(\frac{B-C}{A}\right)= \left(A>0,\theta \ne \left(2n+1\right)\frac{\pi }{2},n\in Z\right)$
 

We have$\frac{\sin \theta }{\cos 3\theta }=\frac{1}{2}\left(\frac{\sin 2\theta }{\cos \theta \cos 3\theta }\right)=\frac{1}{2}\left(\tan 3\theta -\tan \theta \right)$

Similarly $\frac{\sin 3\theta }{\cos 9\theta }=\frac{1}{2}\left(\tan 9\theta -\tan 3\theta \right)$

$\frac{\sin 9\theta }{\cos 27\theta }=\frac{1}{2}\left(\tan 27\theta -\tan 9\theta \right)$ and $\frac{\sin 27\theta }{\cos 81\theta }=\frac{1}{2}\left(\tan 81\theta -\tan 27\theta \right)$

$\therefore \frac{\sin \theta }{\cos 3\theta }+\frac{\sin 3\theta }{\cos 9\theta }+\frac{\sin 9\theta }{\cos 27\theta }+\frac{\sin 27\theta }{\cos 81\theta }=\frac{1}{2}\left(\tan 81\theta -\tan \theta \right)$

$\Rightarrow \frac{B-C}{A}=\frac{81+1}{1/2}=164$