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

Question

¿Cómo simplificar esta expresión?

PS

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

Answer 1

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

Uso de la identidad $$\sec^2(\theta_{i})=4\csc^2(2\theta_{i})-\csc^2(\theta_{i})$$ and using $ \ displaystyle \ theta_ {i +1} = 2 \ theta_ {i}. $

y tenemos $$\csc^2(\theta_{n+1})=\csc^2(\theta_{1}).$ $

PS

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