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

$$\begin{gathered} Wehave\frac{\sin \theta }{\cos 3\theta }=\frac{1}{2}\left(\frac{\sin 2\theta }{\cos \theta \cos 3\theta }\right) \\ \frac{1}{2}\frac{\sin (3\theta -\theta )}{\cos \theta \cos 3\theta }=\frac{1}{2}\frac{\sin 3\theta \cos \theta -\cos 3\theta \sin \theta }{\cos \theta \cos 3\theta } \\ \hspace{0.17em}\frac{\sin \theta }{\cos 3\theta }=\frac{1}{2}\left(\tan 3\theta -\tan \theta \right) -----\left(1\right) \end{gathered}$$

$\begin{array}{l}Similarly\frac{\sin 3\theta }{\cos 9\theta }=\frac{1}{2}\left(\tan 9\theta -\tan 3\theta \right) ----\left(2\right)\\ and\frac{\sin \text{9}\theta }{\cos 27\theta }=\frac{1}{2}\left(\tan 27\theta -\tan 9\theta \right) ----\left(3\right)\end{array}$

adding (1) ,(2) ,(3)

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

$\therefore \left(\frac{B+C}{A}\right)=\frac{27+1}{{\displaystyle \frac{1}{2}}}=56$