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If $\frac{1-\tan \theta}{1+\tan \theta}=\frac{1}{\sqrt{3}}$, where $\theta \in\left(0, \frac{\pi}{2}\right)$, then $\theta=$
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Verified Answer
The correct answer is:
$\frac{\pi}{12}$
Given $\frac{1-\tan \theta}{1+\tan \theta}=\frac{1}{\sqrt{3}}$
$\therefore \tan \left(\frac{\pi}{4}-\theta\right)=\frac{1}{\sqrt{3}}$
Comparing with $\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}$, we write
$$
\begin{aligned}
& \frac{\pi}{4}-\theta=\frac{\pi}{6} \\
\therefore & \theta=\frac{\pi}{4}-\frac{\pi}{6}=\frac{\pi}{12}
\end{aligned}
$$
$\therefore \tan \left(\frac{\pi}{4}-\theta\right)=\frac{1}{\sqrt{3}}$
Comparing with $\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}$, we write
$$
\begin{aligned}
& \frac{\pi}{4}-\theta=\frac{\pi}{6} \\
\therefore & \theta=\frac{\pi}{4}-\frac{\pi}{6}=\frac{\pi}{12}
\end{aligned}
$$
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