$\text{ If }{I}_n={\int }_0^{\pi } {\tan }^{n}xdx\text{ then }$ $I_n+I_{n-2}=\frac{1}{n-1}(n\in N,n\ge 2)$

$I_n={\int }_0^{\pi } {\tan }^{n}xdx={\int }_0^{\frac{\pi }{4}} {\tan }^{n-2}x{\tan }^{2}xdx$

$={\int }_0^{\frac{\pi }{4}} {\tan }^{n-2}x\left({\sec }^{2}x-1\right)dx$

$={\int }_0^{\frac{\pi }{4}} {\tan }^{n-2}x{\sec }^{2}xdx-{\int }_0^{\frac{\pi }{4}} {\tan }^{n-2}xdx$

$I_n={\left[\frac{{\tan }^{n-2+1}x}{{\tan }^{n-1}x}\right]}^{\frac{\pi }{4}}-I_{n-2}$

$I_n={\left[\frac{{\tan }^{n-1}x}{n-1}\right]}_0^{\frac{\pi }{4}}-I_{n-2}$

$I_n+I_{n-2}=\left[\frac{{\tan }_{\frac{\pi }{4}}^{n-1}}{n-1}\right]$

$I_n+I_{n-2}=\frac{1}{n-1}$