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Black Advanced Problems in Mathematics
Exam
Black Advanced Problems in Mathematics
Subject
Mathematics
Chapter
Sequences and Series

Black Advanced Problems in Mathematics · Mathematics · Sequences and Series

If (α - 2β) , α , (α + 2β) are in arithmetical progression then ^2α...

If sec⁡(α−2β)\sec(\alpha - 2\beta), sec⁡α\sec\alpha, sec⁡(α+2β)\sec(\alpha + 2\beta) are in arithmetical progression then cos⁡2α=λcos⁡2β\cos^2\alpha = \lambda \cos^2\beta (β≠nπ,n∈I\beta \ne n\pi, n \in I) the value of λ\lambda is:

Options

  1. A11
  2. B22
  3. C33
  4. D12\frac{1}{2}

Correct answer

C. 33

Step-by-step solution

2sec⁡α=sec⁡(α−2β)+sec⁡(α+2β)2cos⁡α=cos⁡(α+2β)+cos⁡(α−2β)cos⁡(α−2β)cos⁡(α+2β)2cos⁡α=2cos⁡αcos⁡2βcos⁡2α−sin⁡22β2\sec\alpha = \sec(\alpha - 2\beta) + \sec(\alpha + 2\beta)\frac{2}{\cos\alpha} = \frac{\cos(\alpha + 2\beta) + \cos(\alpha - 2\beta)}{\cos(\alpha - 2\beta)\cos(\alpha + 2\beta)}\frac{2}{\cos\alpha} = \frac{2\cos\alpha\cos 2\beta}{\cos^2\alpha - \sin^2 2\beta} Cross-multiplying gives: cos⁡2α−sin⁡22β=cos⁡2αcos⁡2βcos⁡2α(1−cos⁡2β)=sin⁡22β=1−cos⁡22β=(1−cos⁡2β)(1+cos⁡2β)\cos^2\alpha - \sin^2 2\beta = \cos^2\alpha \cos 2\beta\cos^2\alpha (1 - \cos 2\beta) = \sin^2 2\beta = 1 - \cos^2 2\beta = (1 - \cos 2\beta)(1 + \cos 2\beta) Since β≠nπ\beta \ne n\pi, 1−cos⁡2β≠01 - \cos 2\beta \ne 0. cos⁡2α=1+cos⁡2β=2cos⁡2β\cos^2\alpha = 1 + \cos 2\beta = 2\cos^2\beta Thus, λ=2\lambda = 2.
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