IIT JEE , AIEEE Forum - Mathematics , Trignometry

ramkumar_november

let

x = $(1+sec2\theta).(1+sec2^2\theta).(1+sec2^3\theta).........(1+sec2^n \theta)$

x = $\frac{(1+cos2\theta).(1+cos2^2\theta).(1+cos2^3\theta).........(1+cos2^n \theta)}{cos2\theta.cos2^2\theta.cos2^3\theta.....cos2^n\theta}$

x = $\frac{2cos^2\theta.2cos^2 2\theta.2cos^2 2^2\theta.........2cos^2 2^{n-1} \theta}{cos2\theta.cos2^2\theta.cos2^3\theta.....cos2^n\theta}$

cancelling the common terms in numerator and denominator....

x = $\frac{2^ncos^2\theta}{cos2^n\theta}\bigg(cos2\theta.cos2^2\theta.cos2^3\theta.......cos2^{n-1}\theta\bigg)$

x = $\frac{2^ncos^2\theta}{cos2^n\theta}.\frac{sin2^n\theta}{2^nsin\theta cos\theta}$

x = $tan2^n\theta.cot\theta$

so

$(1+sec2\theta).(1+sec2^2\theta).(1+sec2^3\theta).........(1+sec2^n \theta)$ =

$tan2^n\theta.cot\theta$

i have used this formula in my proof.....

$cos\theta.cos2\theta.cos2^2\theta....cos2^{n-1}\theta\;=\;\frac{sin2^n\theta}{2^n sin\theta}$

clear with the proof??