8 =2-1 - Vidyadip

Question

$\tan \frac{\pi}{8}=\sqrt{2}-1$

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Answer

We know that,
$\begin{aligned}
& \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta} \\
& \tan \frac{\pi}{4}=\frac{2 \tan \frac{\pi}{8}}{1-\tan ^2 \frac{\pi}{8}}
\end{aligned}$
Let $\tan \frac{\pi}{8}=\mathrm{t}$
$\begin{array}{ll}
\therefore & \frac{2 \mathrm{t}}{1-\mathrm{t}^2}=1 \\
\therefore & 2 \mathrm{t}=1-\mathrm{t}^2 \\
\therefore & \mathrm{t}^2+2 \mathrm{t}-1=0 \\
\therefore & \mathrm{t}=\frac{-2 \pm \sqrt{4+4}}{2} \\
& =\frac{-2 \pm 2 \sqrt{2}}{2} \\
& =-1 \pm \sqrt{2} \\
& \mathrm{t}=\tan \frac{\pi}{8}>0 \\
\therefore & \tan \frac{\pi}{8}=\sqrt{2}-1
\end{array}$

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