If $I_n=\int\limits_0^{\pi / 4} \tan ^n \theta d \theta$, then $I_8+I_6=$

$\frac{1}{4}$ $\frac{1}{5}$ $\frac{1}{6}$ $\frac{1}{7}$

$\frac{1}{7}$

$I_8+I_6=\int\limits_0^{\pi / 4}\left(\tan ^8 \theta+\tan ^6 \theta\right) d \theta$ $=\int\limits_0^{\pi / 4} \tan ^6 \theta \sec ^2 \theta d \theta=\left[\frac{\tan ^7 \theta}{7}\right]_0^{\pi / 4}=\frac{1}{7}$ Hence (4) is the correct answer.