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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2018 · Mathematics · Trigonometric Ratios & Identities

What is 105^ ^ + 105^ ^ equal to?

What is sin⁡105∘+cos⁡105∘\sin 105^{{}^{\circ}}+\cos 105^{{}^{\circ}} equal to?

Options

  1. Asin⁡50∘\sin 50^{{}^{\circ}}
  2. Bcos⁡50∘\cos 50^{{}^{\circ}}
  3. C12\frac{1}{\sqrt{2}}
  4. D0

Correct answer

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

Step-by-step solution

Sin⁡105∘+cos⁡105∘=sin⁡(60∘+45∘)+cos⁡(60∘+45∘)=(sin⁡60∘⋅cos⁡45∘+cos⁡60∘sin⁡45∘)+(cos⁡60∘cos⁡45∘=32⋅12+12⋅12+12⋅12−32⋅12=222=12\begin{aligned} & \operatorname{Sin} 105^{{}^{\circ}}+\cos 105^{{}^{\circ}} \\ & =\sin \left(60^{{}^{\circ}}+45^{{}^{\circ}}\right)+\cos \left(60^{{}^{\circ}}+45^{{}^{\circ}}\right) \\ & =\left(\sin 60^{{}^{\circ}} \cdot \cos 45^{{}^{\circ}}+\cos 60^{{}^{\circ}} \sin 45^{{}^{\circ}}\right)+\left(\cos 60^{{}^{\circ}} \cos 45^{{}^{\circ}}\right. \\ & =\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}+\frac{1}{2} \cdot \frac{1}{\sqrt{2}}+\frac{1}{2} \cdot \frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} \\ & =\frac{2}{2 \sqrt{2}}=\frac{1}{\sqrt{2}}\end{aligned}
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