If α= log e √ cot (π/12), then find the value of (( displaystyle∑ k =0∞ e -2 k α/ displaystyle∑ k =0∞(-1) k e -2 k α))2

Question

If α= log e √ cot (π/12), then find the value of (( displaystyle∑ k =0∞ e -2 k α/ displaystyle∑ k =0∞(-1) k e -2 k α))2 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Expression (1+ e -2 α+ e -4 α+ ldots ∞/1- e -2 α+ e -4 α+ ldots ∞) =(1+ e -2 α/1- e -2 α)=(( e 2 α+1/ e 2 α-1)) As, e α=√ cot (π/12) ⇒ e 2 α=(2+√3) ∴ d Expression =√3

Q. If $α = lo g_{e} cot \frac{π}{12}$ , then find the value of $⎝ ⎛ \frac{k = 0 \sum \infty e ^{- 2 k α}}{k = 0 \sum \infty ( - 1 ) ^{k} e ^{- 2 k α}} ⎠ ⎞^{2}$

Expression
$\frac{1 + e ^{- 2 α} + e ^{- 4 α} + \dots \infty}{1 - e ^{- 2 α} + e ^{- 4 α} + \dots \infty}$
$= \frac{1 + e ^{- 2 α}}{1 - e ^{- 2 α}} = (\frac{e ^{2 α} + 1}{e ^{2 α} - 1})$
As, $e^{α} = cot \frac{π}{12}$
$\Rightarrow e^{2 α} = (2 + 3)$
$∴ d$ Expression $= 3$

Q. If α=loge​cot12π​​, then find the value of ⎝⎛​k=0∑∞​(−1)ke−2kαk=0∑∞​e−2kα​⎠⎞​2

Expression 1−e−2α+e−4α+…∞1+e−2α+e−4α+…∞​ =1−e−2α1+e−2α​=(e2α−1e2α+1​) As, eα=cot12π​​ ⇒e2α=(2+3​) ∴d Expression =3​

Q. If $α = lo g_{e} cot \frac{π}{12}$ , then find the value of $⎝ ⎛ \frac{k = 0 \sum \infty e ^{- 2 k α}}{k = 0 \sum \infty ( - 1 ) ^{k} e ^{- 2 k α}} ⎠ ⎞^{2}$

Expression
$\frac{1 + e ^{- 2 α} + e ^{- 4 α} + \dots \infty}{1 - e ^{- 2 α} + e ^{- 4 α} + \dots \infty}$
$= \frac{1 + e ^{- 2 α}}{1 - e ^{- 2 α}} = (\frac{e ^{2 α} + 1}{e ^{2 α} - 1})$
As, $e^{α} = cot \frac{π}{12}$
$\Rightarrow e^{2 α} = (2 + 3)$
$∴ d$ Expression $= 3$