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Question
Prove that (1 + sec 2θ)(1 + sec 4θ) ... (1 + sec 2nθ) = tan 2nθ
Solution
L.H.S (1 + sec 2θ) = `1 + 1/(cos2theta) + (cos 2theta + 1)/(cos 2theta)`
= `(2cos^2theta)/(cos 2theta)`
(1 + sec 4θ) = `1 + 1/(cos 4theta)`
= `(cos 4theta + 1)/(cos 4theta)`
= `(2 cos^2 (2theta))/(cos 4theta)`
(1 + sec 2nθ) = `1 + 1/(2^"n" theta)`
= `(cos 2^"n" theta + 1)/(2^"n" theta)`
= `(2 cos^2 2^("n" - 1) theta)/(cos 2^"n" theta)`
(1 + sec 2θ)(1 + sec 4θ) ... (1 + sec 2nθ)
= `(2^"n" cos^2 theta)/(cos 2theta) * (cos^2 2theta)/(cos 4 theta) ... (cos^2 2^("n" - 1) theta)/(cos 2^"n" theta)`
= `(2^"n" cos theta)/(cos 2^"n" theta) {cos theta* cos 2theta ... cos 2^("n" - 1) theta}`
= `(2^"n" costheta{sin 2^"n"theta})/(2^"n" sintheta cos 2^"n" theta)`
= tan 2nθ . cosθ
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