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Prove that (1 + sec 2θ)(1 + sec 4θ) ... (1 + sec 2nθ) = tan 2nθ - Mathematics

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प्रश्न

Prove that (1 + sec 2θ)(1 + sec 4θ) ... (1 + sec 2nθ) = tan 2nθ

योग
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उत्तर

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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Trigonometric Functions and Their Properties
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometry - Exercise 3.5 [पृष्ठ ११८]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 3 Trigonometry
Exercise 3.5 | Q 10 | पृष्ठ ११८

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