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Question Number 28608 by abdo imad last updated on 27/Jan/18
transform tanp−tanq then find the value of   Σ_(k=1) ^n    (1/(cos(kθ)cos((k+1)θ)) .  θ∈R.
transformtanptanqthenfindthevalueofk=1n1cos(kθ)cos((k+1)θ.θR.
Answered by Tinkutara last updated on 28/Jan/18
(1/(cos kθcos (k+1)θ))  =((cos [(k+1)θ−kθ])/(cos kθcos (k+1)θcos θ))  =((cos (k+1)θcos θ+sin (k+1)θsin θ)/(cos kθcos (k+1)θcos θ))  =((1+tan (k+1)θtan θ)/(cos θ))  =((tan (k+1)θ−tan θ)/(sin θ))  Telescoping it becomes  Σ_(k=1) ^n ((tan (k+1)θ−tan θ)/(sin θ))  =((tan (n+1)θ−tan θ)/(sin θ))
1coskθcos(k+1)θ=cos[(k+1)θkθ]coskθcos(k+1)θcosθ=cos(k+1)θcosθ+sin(k+1)θsinθcoskθcos(k+1)θcosθ=1+tan(k+1)θtanθcosθ=tan(k+1)θtanθsinθTelescopingitbecomesnk=1tan(k+1)θtanθsinθ=tan(n+1)θtanθsinθ

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