The absolute value of the expression $\tan \frac{\mathrm{\pi }}{16}+\tan \frac{5\mathrm{\pi }}{16}$$+\tan \frac{9\mathrm{\pi }}{16}+\tan \frac{13\mathrm{\pi }}{16}$ is

$\text{ Let }\mathrm{\theta }=\frac{\mathrm{\pi }}{16}\text{ or }8\mathrm{\theta }=\frac{\mathrm{\pi }}{2}$

$\begin{array}{l}\mathrm{y}=\tan \mathrm{\theta }+\tan 5\mathrm{\theta }+\tan 9\mathrm{\theta }+\tan 13\mathrm{\theta }\\ \therefore \mathrm{y}=(\tan \mathrm{\theta }-\cot \mathrm{\theta })+(\tan 5\mathrm{\theta }-\cot 5\mathrm{\theta })\\ [\text{ As }\tan 13\mathrm{\theta }=\tan (8\mathrm{\theta }+5\mathrm{\theta })=-\cot 5\mathrm{\theta }\text{ and }\tan 9\mathrm{\theta }=\tan (8\mathrm{\theta }+\mathrm{\theta })=-\cot \mathrm{\theta }]\\ =(\tan \mathrm{\theta }-\cot \mathrm{\theta })+(\cot 3\mathrm{\theta }-\tan 3\mathrm{\theta })\\ =\frac{{\sin }^2\mathrm{\theta }-{\cos }^2\mathrm{\theta }}{\sin \mathrm{\theta cos}\mathrm{\theta }}+\frac{{\cos }^{2}3\mathrm{\theta }-{\sin }^{2}3\mathrm{\theta }}{\sin 3\mathrm{\theta cos}3\mathrm{\theta }}\end{array}$

$\begin{array}{l}\Rightarrow \mathrm{y}=2\left[\frac{\cos 6\mathrm{\theta }}{\sin 6\mathrm{\theta }}-\frac{\cos 2\mathrm{\theta }}{\sin 2\mathrm{\theta }}\right]\\ =2\left[\frac{\sin 20\cos 6\mathrm{\theta }-\cos 2\mathrm{\theta sin}6\mathrm{\theta }}{\sin 6\mathrm{\theta sin}2\mathrm{\theta }}\right]\\ =-2\left[\frac{\sin 4\mathrm{\theta }}{\cos 2\mathrm{\theta sin}2\mathrm{\theta }}\right]=-4\left(\because 6\mathrm{\theta }=\frac{\mathrm{\pi }}{2}-2\mathrm{\theta }\right)\end{array}$

Hence, absolute value is 4.