Simplifying $\frac{{\sum_{i=1}^{i=n}}1+\tan^2\theta_i}{{\sum_{i=1}^{i=n}}1+\cot^2\theta_i}$, where $\theta_i = \frac{2^{i-1}\pi}{2^n+1}$

Question

How to simplify this expression?

$$\frac{\sum_{i=1}^{n}\left(1+\tan^2\theta_i\right)}{\sum_{i=1}^{n}\left(1+\cot^2\theta_i\right)}$$

where $$\theta_i = \frac{2^{i-1}\pi}{2^n+1}$$

Answer 1

Given $\displaystyle \theta_{i}=\frac{2^{i-1}}{2^n+1}$

Using the Identity $$\sec^2(\theta_{i})=4\csc^2(2\theta_{i})-\csc^2(\theta_{i})$$ and using $\displaystyle \theta_{i+1}=2\theta_{i}.$

and we have $$\csc^2(\theta_{n+1})=\csc^2(\theta_{1}).$$

$$\displaystyle \sum^{n}_{i=1}\sec^2(\theta_{i})=4\sum^{n}_{i=1}\csc^2(\theta_{i+1})-\sum^{n}_{i=1}$$

$$\csc^2(\theta_{i})=3\sum^{n}_{i=1}\csc^2(\theta_{i}).$$