Given integral is,
$I_{n} =\int\limits_{0}^{\pi / 1} \tan ^{n} x d x$
$I_{\vec{n}} =\int\limits_{0}^{\pi / 4} \tan ^{n-2} x \cdot \tan ^{2} x d x$
$=\int\limits_{0}^{\pi 4} \tan ^{n-2} x \cdot\left(\sec ^{2} x-1\right) d x$
$I_{n}=\int\limits_{0}^{\pi / 4} \tan ^{n-2} x \cdot \sec ^{2} x d x$
$-\int\limits_{0}^{\pi / 4} \tan ^{n-2} x d x$
$I_{n}=\int\limits_{0}^{1} t^{n-2} d t-\int\limits_{0}^{\pi / 4} \tan ^{n-2} x d x$
$I_{n}=\left[\frac{t^{n-1}}{n-1}\right]_{0}^{1}-I_{n-2}$
$\left\{\text { put } \tan x=t \\ \Rightarrow \sec ^{2} x d x=d t\right\}$
$I_{n}+I_{n-2}=\frac{1}{n-1}$
Put $n=10$, we get
$I_{10}+I_{8}=\frac{1}{9}$