# If $P=\tan(3^{n+1}\theta)-\tan\theta$ and $Q=\sum_{r=0}^n\frac{\sin(3^r\theta)}{\cos(3^{r+1}\theta)}$,

If $P=\tan(3^{n+1}\theta)-\tan\theta$ and $Q=\sum_{r=0}^n\frac{\sin(3^r\theta)}{\cos(3^{r+1}\theta)}$,then relate $P$ and $Q$

$Q=\sum_{r=0}^n\frac{\sin(3^r\theta)}{\cos(3^{r+1}\theta)}$
$=\sum_{r=0}^n\frac{\sin(3^r\theta)}{\cos(3.3^{r}\theta)}=\sum_{r=0}^n\frac{\sin(3^r\theta)}{4\cos^3(3^{r}\theta)-3\cos(3^{r}\theta)}$

I do not know how to change into telescoping series.

## 1 Answer

HINT:

$$\tan3A-\tan A=\dfrac{\sin(3A-A)}{\cos3A\cos A}=\dfrac{2\sin A}{\cos3A}$$

Do you recognize the Telescoping nature?