If $$b_n = \int_0^{\pi/2} \frac{\cos^2(nx)}{\sin x} dx, n \in \mathbb{N}$$, then

We are given $$b_n = \int_0^{\pi/2} \frac{\cos^2(nx)}{\sin x}\,dx$$ for $$n \in \mathbb{N}$$.

Using the identity $$\cos^2(nx) = \frac{1 + \cos(2nx)}{2}$$, we get:
$$b_n = \frac{1}{2}\int_0^{\pi/2} \frac{1 + \cos(2nx)}{\sin x}\,dx = \frac{1}{2}\int_0^{\pi/2} \frac{1}{\sin x}\,dx + \frac{1}{2}\int_0^{\pi/2} \frac{\cos(2nx)}{\sin x}\,dx$$.

Let us compute $$b_{n+1} - b_n$$. We have:
$$b_{n+1} - b_n = \frac{1}{2}\int_0^{\pi/2} \frac{\cos(2(n+1)x) - \cos(2nx)}{\sin x}\,dx$$.

Using the identity $$\cos A - \cos B = -2\sin\Bigl(\frac{A+B}{2}\Bigr)\sin\Bigl(\frac{A-B}{2}\Bigr)$$, we obtain $$\cos(2(n+1)x) - \cos(2nx) = -2\sin((2n+1)x)\sin(x)$$. Hence
$$b_{n+1} - b_n = \frac{1}{2}\int_0^{\pi/2} \frac{-2\sin((2n+1)x)\sin x}{\sin x}\,dx = -\int_0^{\pi/2} \sin((2n+1)x)\,dx$$.

Evaluating the integral gives:
$$-\int_0^{\pi/2} \sin((2n+1)x)\,dx = \frac{\cos((2n+1)x)}{2n+1}\Big|_0^{\pi/2} = \frac{\cos\left(\frac{(2n+1)\pi}{2}\right) - 1}{2n+1}$$.

Since $$(2n+1)$$ is odd, $$\cos\left(\frac{(2n+1)\pi}{2}\right) = 0$$. Therefore
$$b_{n+1} - b_n = \frac{-1}{2n+1}$$.

Computing the required differences, we have $$b_3 - b_2 = \frac{-1}{5}$$, $$b_4 - b_3 = \frac{-1}{7}$$, $$b_5 - b_4 = \frac{-1}{9}$$.

Taking reciprocals yields $$\frac{1}{b_3 - b_2} = -5$$, $$\frac{1}{b_4 - b_3} = -7$$, $$\frac{1}{b_5 - b_4} = -9$$, which are in A.P. with common difference $$-7 - (-5) = -2$$.

The answer is Option D: $$\frac{1}{b_3 - b_2}, \frac{1}{b_4 - b_3}, \frac{1}{b_5 - b_4}$$ are in A.P. with common difference $$-2$$.