Question

Find the value of $\tan \frac{\pi }{8}.$

Answer 1

Let $\theta =\frac{\pi }{8}$ then $2\theta =\frac{\pi }{4}$
$\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$
$\Rightarrow \tan \frac{\pi }{4}=\frac{2\tan \frac{\pi }{8}}{1-{\tan }^2\frac{\pi }{8}}\Rightarrow 1=\frac{2\tan \frac{\pi }{8}}{1-{\tan }^2\frac{\pi }{8}}$

Let $y=\tan \frac{\pi }{8}$
$\Rightarrow 1=\frac{2y}{1-y^2}\Rightarrow 1-y^2=2y$
$\Rightarrow y^2+2y-1=0$
The above equation is of the form $a{x}^2+bx+c=0$
$a=1,b=2,c=-1$
$\therefore y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$\Rightarrow y=\frac{-2\pm \sqrt{(2{)}^2-4(1\left)\right(-1)}}{2\times 1}\Rightarrow y=\frac{-2\pm \sqrt{4+4}}{2\times 1}$
$\Rightarrow y=\frac{-2\pm 2\sqrt{2}}{2}\Rightarrow y=-1\pm \sqrt{2}$
$\Rightarrow \tan \frac{\pi }{8}=-1\pm \sqrt{2}$

But $\tan \frac{\pi }{8}=-1-\sqrt{2}$ is not possible as $\frac{\pi }{8}$ lies in the first quadrant and $\tan$ is positive in first quadrant.
$\therefore \tan \frac{\pi }{8}=-1+\sqrt{2}=\sqrt{2}-1$