For a positive integer n, let ${\mathrm{f}}_{\mathrm{n}}\left(\mathrm{\theta }\right)=\frac{\tan \mathrm{\theta }}{2}(1+\sec \mathrm{\theta })(1+\sec 2\mathrm{\theta })\dots \left(1+\sec 2^{\mathrm{n}}\mathrm{\theta }\right)$ then

$\begin{array}{l}\because \left(\tan \frac{\mathrm{\theta }}{2}\right)(1+\sec \mathrm{\theta })=\tan \left(\frac{\mathrm{\theta }}{2}\right)\left(\frac{1+\cos \mathrm{\theta }}{\cos \mathrm{\theta }}\right)\\ =\frac{\sin \frac{\mathrm{\theta }}{2}}{\cos \frac{\mathrm{\theta }}{2}}\cdot \frac{2{\cos }^2\frac{\mathrm{\theta }}{2}}{\cos \mathrm{\theta }}=\frac{2\sin \frac{\mathrm{\theta }}{2}\cdot \cos \frac{\mathrm{\theta }}{2}}{\cos \mathrm{\theta }}\\ =\frac{\sin \mathrm{\theta }}{\cos \mathrm{\theta }}=\tan \mathrm{\theta }----\left(\mathrm{i}\right)\end{array}$
By repeated use of Eq. (i), we have
$\begin{array}{c}{\mathrm{f}}_{\mathrm{n}}\left(\mathrm{\theta }\right)=\tan \mathrm{\theta }(1+\sec 2\mathrm{\theta })(1+\sec 4\mathrm{\theta })\dots \left(1+\sec 2^{\mathrm{n}}\mathrm{\theta }\right)\\ =\tan 2\mathrm{\theta }(1+\sec 4\mathrm{\theta })\dots \left(1+\sec 2^{\mathrm{n}}\mathrm{\theta }\right)\end{array}$
$\begin{array}{l}=\tan 4\mathrm{\theta }(1+\sec 8\mathrm{\theta })\dots \left(1+\sec 2^{\mathrm{n}}\mathrm{\theta }\right)=\dots .=\tan 2^{\mathrm{n}}\mathrm{\theta }\\ \\ \mathrm{Now} ,{\mathrm{f}}_2\left(\frac{\mathrm{\pi }}{16}\right)=\tan \left(2^2\frac{\mathrm{\pi }}{16}\right)=\tan \frac{\mathrm{\pi }}{4}=1\\ {\mathrm{f}}_3\left(\frac{\mathrm{\pi }}{32}\right)=\tan \left(2^3\frac{\mathrm{\pi }}{32}\right)=\tan \frac{\mathrm{\pi }}{4}=1\\ {\mathrm{f}}_4\left(\frac{\mathrm{\pi }}{64}\right)=\tan \left(2^4\frac{\mathrm{\pi }}{64}\right)=\tan \frac{\mathrm{\pi }}{4}=1\\ \mathrm{and} {\mathrm{f}}_5\left(\frac{\mathrm{\pi }}{128}\right)=\tan \frac{\mathrm{\pi }}{4}=1\end{array}$