$$\eqalign{
& \tan 2x = {{2\tan x} \over {1 - {{\tan }^2}x}} \cr
& \Rightarrow \tan {\pi \over 4} = {{2\tan {\pi \over 8}} \over {1 - {{\tan }^2}{\pi \over 8}}} = 1 \cr
& \Leftrightarrow 2\tan {\pi \over 8} = 1 - {\tan ^2}{\pi \over 8} \cr
& \Leftrightarrow {\tan ^2}{\pi \over 8} + 2\tan {\pi \over 8} - 1 = 0 \cr
& Dat\,\,t = \tan {\pi \over 8} \cr
& \Rightarrow {t^2} + 2t - 1 = 0 \Leftrightarrow t = - 1 \pm \sqrt 2 \cr
& \Rightarrow \tan {\pi \over 8} = - 1 \pm \sqrt 2 \cr
& Do\,\,\left\{ \matrix{
\cos {\pi \over 8} > 0 \hfill \cr
\sin {\pi \over 8} > 0 \hfill \cr} \right. \Rightarrow \tan {\pi \over 8} > 0 \cr
& \Rightarrow \tan {\pi \over 8} = \sqrt 2 - 1 \cr} $$
