Let $$\vec{a_k} = (\tan\theta_k)\hat{i} + \hat{j}$$ and $$\vec{b_k} = \hat{i} - (\cot\theta_k)\hat{j}$$, where $$\theta_k = \frac{2^{k-1}\pi}{2^n + 1}$$, for some $$n \in \mathbb{N}$$, $$n > 5$$. Then the value of $$\frac{\sum_{k=1}^{n}|\vec{a_k}|^2}{\sum_{k=1}^{n}|\vec{b_k}|^2}$$ is _____.

$$|\vec{a_k}|^2 = \tan^2\theta_k + 1 = \sec^2\theta_k$$

$$|\vec{b_k}|^2 = 1 + \cot^2\theta_k = \csc^2\theta_k$$

$$S = \frac{\sum_{k=1}^{n}|\vec{a_k}|^2}{\sum_{k=1}^{n}|\vec{b_k}|^2} = \frac{\sum_{k=1}^{n} \sec^2\theta_k}{\sum_{k=1}^{n} \csc^2\theta_k}$$

Given $$\theta_k = \frac{2^{k-1}\pi}{2^n + 1}$$: $$\sum_{k=1}^{n} \tan^2\theta_k = \sum_{k=1}^{n} \cot^2\theta_k$$

$$\sum_{k=1}^{n} \sec^2\theta_k = \sum_{k=1}^{n} (1 + \tan^2\theta_k) = n + \sum_{k=1}^{n} \tan^2\theta_k$$

$$\sum_{k=1}^{n} \csc^2\theta_k = \sum_{k=1}^{n} (1 + \cot^2\theta_k) = n + \sum_{k=1}^{n} \cot^2\theta_k$$

Hence, $$\sum_{k=1}^{n} \sec^2\theta_k = \sum_{k=1}^{n} \csc^2\theta_k$$

$$S = \frac{\sum_{k=1}^{n} \sec^2\theta_k}{\sum_{k=1}^{n} \sec^2\theta_k} = 1$$