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

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

• what ? how? could you please recheck your calculations? – maveric Feb 13 at 17:28
• seriously, why is this question being downvoted? – maveric Feb 13 at 17:33
• Can you show some of your efforts? This will help us pick up where you left off :) – YiFan Feb 13 at 22:52
• i tried with roots of unity. with n=3, formed an equation whose roots are square the given angles, but one angle tan 3pi/9 was also root. but its value was known so i could find. but then not able to generalize for n. – maveric Feb 14 at 3:46
• Getting answer $3....$ – jacky Feb 14 at 11:26

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 $$\csc^2(\theta_{n+1})=\csc^2(\theta_{1}).$$
$$\displaystyle \sum^{n}_{i=1}\sec^2(\theta_{i})=4\sum^{n}_{i=1}\csc^2(\theta_{i+1})-\sum^{n}_{i=1}$$
$$\csc^2(\theta_{i})=3\sum^{n}_{i=1}\csc^2(\theta_{i}).$$