$\left|a_k\right|^2 = 1+\tan^2\theta_k = \sec^2\theta_k$

$\left|b_k\right|^2 = 1+\cot^2\theta_k = \csc^2\theta_k$

Since, $\cot\theta-\tan\theta = 2\cot2\theta$

$\therefore\; -\csc^2\theta-\sec^2\theta = -4\csc^2 2\theta$

$\sec^2\theta = 4\csc^2 2\theta - \csc^2\theta$

$\sum \sec^2\theta_k = 4\sum \csc^2 2\theta_k - \sum \csc^2\theta_k$

Now

$\sum \csc^2 2\theta_k = \sum \csc^2\theta_k$

since

$\displaystyle \sum_{k=1}^{n} \csc^2\!\left( \dfrac{2^k\pi}{2^n+1} \right) = \sum_{k=1}^{n} \csc^2\!\left( \dfrac{2^{k-1}\pi}{2^n+1} \right)$

because $\displaystyle \csc\!\left( \dfrac{2^n\pi}{2^n+1} \right) = \csc\!\left( \dfrac{\pi}{2^n+1} \right)$

$\therefore\; \sum \sec^2\theta_k = 3\sum \csc^2\theta_k$

$\displaystyle \frac{\sum \sec^2\theta_k} {\sum \csc^2\theta_k} = 3$