Question:

<div><p>$$I_n = \int_{0}^{\pi/4} (\tan x)^{n-2} \cdot \tan^2 x \, dx = \int_{0}^{\pi/4} (\tan x)^{n-2} (\sec^2 x - 1) dx$$</p><p>$$= \int_{0}^{\pi/4} (\tan x)^{n-2} \sec^2 x \, dx - \int_{0}^{\pi/4} (\tan x)^{n-2} \, dx = \left[\frac{(\tan x)^{n-1}}{n-1}\right]_0^{\pi/4} - I_{n-2}$$</p><p>$$I_n = \frac{1}{n-1} - I_{n-2} \quad \therefore \quad I_n + I_{n-2} = \frac{1}{n-1}$$</p></div>