Answer
The length of the curve is:
$L\approx 1.2789$
Work Step by Step
The length of a curve with polar equation of one loop of the curve
$$
r=\tan(\theta), \quad\quad \frac{\pi}{6} \leq \theta \leq \frac{\pi}{3}
$$
is given by the following:
$$
\begin{split}
L=& \int_{\pi / 6}^{\pi / 3} \sqrt{r^{2}+\left(\frac{d r}{d \theta}\right)^{2}} d \theta\\
&=\int_{\pi / 6}^{\pi / 3} \sqrt{\tan^{2} \theta+( \sec ^{2} \theta)^{2}} d \theta\\
&=\int_{\pi / 6}^{\pi / 3} \sqrt{\tan^{2} \theta+ \sec ^{4} \theta} d \theta\\
&= \int_{-\pi / 4}^{\pi / 4} \sqrt{1+3 \sin ^{2} 2 \theta} d \theta \\
&\approx 1.2789
\end{split}
$$
