Answer
$$1 + \sqrt 2 $$
Work Step by Step
$$\eqalign{
& \tan \left( {\frac{{3\pi }}{8}} \right) \cr
& {\text{Rewrite }}\frac{{3\pi }}{8}{\text{ as }}\frac{1}{2}\left( {\frac{{3\pi }}{4}} \right),{\text{ then}} \cr
& \tan \left( {\frac{{3\pi }}{8}} \right) = \tan \left[ {\frac{1}{2}\left( {\frac{{3\pi }}{4}} \right)} \right] \cr
& {\text{Where tan}}\frac{\theta }{2} = \frac{{1 - \cos \theta }}{{\sin \theta }},{\text{ let }}\theta = \frac{{3\pi }}{4} \cr
& \tan \left[ {\frac{1}{2}\left( {\frac{{3\pi }}{4}} \right)} \right] = \frac{{1 - \cos \left( {\frac{{3\pi }}{4}} \right)}}{{\sin \left( {\frac{{3\pi }}{4}} \right)}} \cr
& {\text{Where }}\cos \left( {\frac{{3\pi }}{4}} \right) = - \frac{{\sqrt 2 }}{2}{\text{ and }}\sin \left( {\frac{{3\pi }}{4}} \right) =\frac{{\sqrt 2 }}{2} \cr
& \tan \left( {\frac{{3\pi }}{8}} \right) = \frac{{1 - \left( { - \frac{{\sqrt 2 }}{2}} \right)}}{{\frac{{\sqrt 2 }}{2}}} \cr
& {\text{Simplifying}} \cr
& \tan \left( {\frac{{3\pi }}{8}} \right) = \frac{{2 - 2\left( { - \frac{{\sqrt 2 }}{2}} \right)}}{{2\left( {\frac{{\sqrt 2 }}{2}} \right)}} = \frac{{2 + \sqrt 2 }}{{\sqrt 2 }} \cr
& \tan \left( {\frac{{3\pi }}{8}} \right) = \frac{2}{{\sqrt 2 }} + 1 \cr
& \tan \left( {\frac{{3\pi }}{8}} \right) = 1 + \sqrt 2 \cr} $$
