The value of  $\tan \frac{\pi }{16}+2\tan \frac{\pi }{8}+4$ is

$\tan \theta =\cot \theta -2\cot 2\theta$

$=\cot \frac{\pi }{16}-2\cot \frac{\pi }{8}+2(\cot \frac{\pi }{8}-2\cot \frac{\pi }{4})+4$

$=\cot \frac{\pi }{16}-2\cot \frac{\pi }{8}+2\cot \frac{\pi }{8}-4\cot \frac{\pi }{4}+4$

$=\cot \frac{\pi }{16}-4\left(1\right)+4$

$=\cot \frac{\pi }{16}$