Let $f_n\left(\theta \right)=\sum _{r=1}^n\tan \left(\frac{\theta }{2^r}\right)\sec \left(\frac{\theta }{2^{r-1}}\right)$ then which of the following alternative(s) is/are correct?

We have $T_n=\tan \left(\frac{\theta }{2^n}\right)\sec \left(\frac{\theta }{2^{n-1}}\right)=\frac{\sin \left(\frac{\theta }{2^n}\right)}{\cos \left(\frac{\theta }{2^n}\right)\cos \left(\frac{\theta }{2^{n-1}}\right)}=\frac{\sin (\frac{1}{2^{n-1}}-\frac{1}{2^n})\theta }{\cos \left(\frac{\theta }{2^n}\right)\cos \left(\frac{\theta }{2^{n-1}}\right)}$

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

$\therefore f_n\left(\theta \right)=\sum _{n=1}^n(\tan \left(\frac{\theta }{2^{n-1}}\right)-\tan \left(\frac{\theta }{2^n}\right))=(\tan \theta -\tan \frac{\theta }{2^n})$

Now, $f_3\left(2\pi \right)=\tan 2\pi -\tan \left(\frac{2\pi }{8}\right)=0-\tan \frac{\pi }{4}=-1$

$f_4\left(\frac{4\pi }{3}\right)=\tan \frac{4\pi }{3}-\tan \left(\frac{4\pi }{3\times 16}\right)=\sqrt{3}-\tan \frac{\pi }{12}=\sqrt{3}-(2-\sqrt{3})=2(\sqrt{3}-1)$

$f_5\left(4\pi \right)=\tan 4\pi -\tan \left(\frac{4\pi }{32}\right)=0-\tan \frac{\pi }{8}=-(\sqrt{2}-1)=(1-\sqrt{2})$

$f_6\left(48\pi \right)=\tan 48\pi -\tan \left(\frac{48\pi }{64}\right)=0-\tan \left(\frac{3\pi }{4}\right)=-(-1)=1$

But in last 2 cases, we get a term of $\tan \left(\frac{\pi }{2}\right)$  which makes it not defined