If $p=\tan \left(3^{n+1}\theta \right)-\tan \theta andQ=\sum _{ }^{ }\frac{\sin \left(3^r\theta \right)}{\cos \left(3^{r+1}\theta \right)}$, then

$Q=\sum _{r=0}^n\frac{\sin \left(3^r\theta \right)}{\cos \left(3^{r+1}\theta \right)}$

$=\frac{\sin \theta }{\cos 3\theta }+\frac{\sin 3\theta }{\cos 9\theta }+\frac{\sin 9\theta }{\cos 27\theta }+\cdots +\frac{\sin \left(3^n\theta \right)}{\cos \left(3^{n+1}\theta \right)}$

As, $\frac{\sin \theta }{\cos 3\theta }=\frac{2\sin \theta \cdot \cos \theta }{2\cos 3\theta \cdot \cos \theta }=\frac{1}{2}\left[\frac{\sin 2\theta }{\cos 3\theta \cdot \cos \theta }\right]$

$=\frac{1}{2}\left[\frac{\sin (3\theta -\theta )}{\cos 3\theta \cdot \cos \theta }\right]=\frac{1}{2}[\tan 3\theta -\tan \theta ]$

$\therefore Q=\frac{1}{2}[(\tan 3\theta -\tan \theta )+(\tan 9\theta -\tan 3\theta )+\cdots +(\tan \left(3^{n+1}\theta \right)-\tan \left(3^n\theta \right))]$

$=\frac{1}{2}[\tan \left(3^{n+1}\theta \right)-\tan \theta ]$

$Q=\frac{P}{2}\Rightarrow P=2Q$.