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NDA 2020
Exam
NDA
Year
2020
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2020 · Mathematics · Trigonometric Ratios & Identities

Paragraph: Read the following information and answer: Let α=β=15^ ^...

Paragraph: Read the following information and answer: Let α=β=15∘\alpha=\beta=15^{{}^{\circ}}.
Question: What is the value of sin⁡α+cos⁡β\sin \alpha+\cos \beta ?

Options

  1. A12\frac{1}{\sqrt{2}}
  2. B122\frac{1}{2 \sqrt{2}}
  3. C322\frac{\sqrt{3}}{2 \sqrt{2}}
  4. D32\frac{\sqrt{3}}{\sqrt{2}}

Correct answer

D. 32\frac{\sqrt{3}}{\sqrt{2}}

Step-by-step solution

α=β=15∘\alpha=\beta=15^{{}^{\circ}} (given) Now, sin⁡α+cos⁡β\sin \alpha+\cos \beta=2⋅(12⋅sin⁡α+12⋅cos⁡α){∵α=β}=2(sin⁡α⋅cos⁡45∘+cos⁡α⋅sin⁡45∘)=2(sin⁡(α+45∘))=2⋅sin⁡(15∘+45∘)=2⋅sin⁡60∘=2×32=32\begin{aligned} & =\sqrt{2} \cdot\left(\frac{1}{\sqrt{2}} \cdot \sin \alpha+\frac{1}{\sqrt{2}} \cdot \cos \alpha\right) \quad\{\because \alpha=\beta\} \\ & =\sqrt{2}\left(\sin \alpha \cdot \cos 45^{{}^{\circ}}+\cos \alpha \cdot \sin 45^{{}^{\circ}}\right) \\ & =\sqrt{2}\left(\sin \left(\alpha+45^{{}^{\circ}}\right)\right)=\sqrt{2} \cdot \sin \left(15^{{}^{\circ}}+45^{{}^{\circ}}\right) \\ & =\sqrt{2} \cdot \sin 60^{{}^{\circ}}=\sqrt{2} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{\sqrt{2}} \end{aligned}$$
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