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

Answer

Suppose, $x =$ $\frac { \pi } { 8 }$
Then, $2x =$ $\frac { \pi } { 4 }$
Now, $ tan 2x =$ tan $\frac { \pi } { 4 }$
$\Rightarrow \quad \frac { 2 \tan x } { 1 - \tan ^ { 2 } x } = \tan \frac { \pi } { 4 }$ [$\because$$ tan 2x =$ $\frac { 2 \tan x } { 1 - \tan ^ { 2 } x }$]
Putting x = $\frac { \pi } { 8 }$, 
$\Rightarrow \frac { 2 \tan \frac { \pi } { 8 } } { 1 - \tan ^ { 2 } \frac { \pi } { 8 } }$$ = 1$
Let y = tan $\frac { \pi } { 8 }$
Then, $\frac { 2 y } { 1 - y ^ { 2 } }$ = 1
$\Rightarrow$ $2y = 1 - y^2$
$\Rightarrow$ $y^2 + 2y - 1 = 0$
$\therefore$ y = $\frac { - 2 \pm \sqrt { ( 2 ) ^ { 2 } + 4 } } { 2 } = \frac { - 2 \pm \sqrt { 8 } } { 2 } = \frac { - 2 \pm 2 \sqrt { 2 } } { 2 }$
$\Rightarrow$ y = - 1 $\pm \sqrt { 2 }$
But $\frac { \pi } { 8 }$ lies in I quadrant, so y  = tan $\frac { \pi } { 8 }$ is positive.
Hence, tan $\frac { \pi } { 8 }$ = - 1 + $\sqrt { 2 }$

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