If $$I_n = \int_{\pi/4}^{\pi/2} \cot^n x \, dx$$, then

We have $$I_n = \int_{\pi/4}^{\pi/2} \cot^n x\,dx$$. Consider $$I_n + I_{n+2}$$:

$$I_n + I_{n+2} = \int_{\pi/4}^{\pi/2} \cot^n x (1 + \cot^2 x)\,dx = \int_{\pi/4}^{\pi/2} \cot^n x \cdot \csc^2 x\,dx$$.

Using the substitution $$u = \cot x$$, $$du = -\csc^2 x\,dx$$. When $$x = \pi/4$$, $$u = 1$$; when $$x = \pi/2$$, $$u = 0$$. So $$I_n + I_{n+2} = \int_1^0 u^n(-du) = \int_0^1 u^n\,du = \frac{1}{n+1}$$.

Therefore: $$I_2 + I_4 = \frac{1}{3}$$, $$I_3 + I_5 = \frac{1}{4}$$, and $$I_4 + I_6 = \frac{1}{5}$$.

Taking reciprocals: $$\frac{1}{I_2 + I_4} = 3$$, $$\frac{1}{I_3 + I_5} = 4$$, $$\frac{1}{I_4 + I_6} = 5$$.

Since $$3, 4, 5$$ are in arithmetic progression (common difference 1), $$\frac{1}{I_2+I_4}, \frac{1}{I_3+I_5}, \frac{1}{I_4+I_6}$$ are in A.P.