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}

Clarence Hines

Clarence Hines

Answered question

2022-01-25

Simplifying i=1i=n1+tan2θii=1i=n1+cot2θi, where θi=2i1π2n+1

Answer & Explanation

pacetfv

pacetfv

Beginner2022-01-26Added 9 answers

Given θi=2i12n+1
Using the Identity
sec2(θi)=4csc2(2θi)csc2(θi)
and using θi+1=2θi
and we have
csc2(θn+1)=csc2(θ1)
{i=1}nsec2(θi)=4{i=1}ncsc2(θi+1){i=1}n
csc2(θi)=3{i=1}ncsc2(θi)

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