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NDA 2023
Exam
NDA
Year
2023
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2023 · Mathematics · Trigonometric Ratios & Identities

If α=(1)/(7), β= 1 10 ; 0 < α, β < (π)/(2) then what is the value...

If tan⁡α=17,sin⁡β=110;0<α,β<π2\tan \alpha=\frac{1}{7}, \sin \beta=\frac{1}{\sqrt{10}} ; 0 < \alpha, \beta < \frac{\pi}{2} then what is the value of cos⁡(α+2β)\cos (\alpha+2 \beta) ?

Options

  1. A−12-\frac{1}{2}
  2. B−12-\frac{1}{\sqrt{2}}
  3. C12\frac{1}{\sqrt{2}}
  4. D12\frac{1}{2}

Correct answer

C. 12\frac{1}{\sqrt{2}}

Step-by-step solution

 Since, cos⁡(α+2β)=cos⁡αcos⁡2β−sin⁡αsin⁡2β=cos⁡α(1−2sin⁡2β)−sin⁡α(2sin⁡βcos⁡β)=750(1−2×110)−(2×110×310)150=750×810−61050=501050=12\begin{aligned} & \text{ Since, } \cos (\alpha+2 \beta) \\ &=\cos \alpha \cos 2 \beta-\sin \alpha \sin 2 \beta \\ &=\cos \alpha\left(1-2 \sin ^2 \beta\right)-\sin \alpha(2 \sin \beta \cos \beta) \\ &=\frac{7}{\sqrt{50}}\left(1-2 \times \frac{1}{10}\right)-\left(2 \times \frac{1}{\sqrt{10}} \times \frac{3}{\sqrt{10}}\right) \frac{1}{\sqrt{50}} \\ &=\frac{7}{\sqrt{50}} \times \frac{8}{10}-\frac{6}{10 \sqrt{50}}=\frac{50}{10 \sqrt{50}}=\frac{1}{\sqrt{2}}\end{aligned}
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