The value of \(\tan \frac{\pi}{8}\) is

Since \(\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta} \)
\( \therefore \tan \frac{\pi}{4}=\frac{2 \tan \frac{\pi}{8}}{1-\tan ^2 \frac{\pi}{8}} \)
\( \Rightarrow 1=\frac{2 \tan \frac{\pi}{8}}{1-\tan ^2 \frac{\pi}{8}} \)
\( \text { Let } y=\tan \frac{\pi}{8} \)
\( \Rightarrow 1=\frac{2 y}{1-y^2} \)
\( \Rightarrow y^2+2 y-1=0 \)
\( \Rightarrow y=\frac{-2 \pm \sqrt{4+4}}{2} \)
\( \Rightarrow y=\frac{-2 \pm 2 \sqrt{2}}{2} \)
\( \Rightarrow y=-1 \pm \sqrt{2} \)
\( \tan \frac{\pi}{8}=-1 \pm \sqrt{2}\)
Since \(\frac{\pi}{8}\) lies in \(1^{\text {st }}\) quadrant.
\(\therefore \tan \frac{\pi}{8} \neq-1-\sqrt{2}\)
\(\therefore \tan \frac{\pi}{8} =-1+\sqrt{2}\)
\(=\sqrt{2}-1\)