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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2024 · Mathematics · Trigonometric Ratios & Identities

ABC is a triangle such that angle C=60^ ^ , then what is A+ B ( A-B...

ABC is a triangle such that angle C=60∘\mathrm{C}=60^{{}^{\circ}}, then what is \frac{\cos \mathrm{A}+\cos \mathrm{B}}{\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)} equal to?

Options

  1. A2
  2. B2\sqrt{2}
  3. C1
  4. D12\frac{1}{\sqrt{2}}

Correct answer

C. 1

Step-by-step solution

cos⁡A+cos⁡Bcos⁡(A−B2)=2cos⁡A+B2⋅cos⁡A−B2cos⁡(A−B2)=2cos⁡(π−C2)=2cos⁡(π2−C2)=2sin⁡C2=2sin⁡30∘=2×12=1.\begin{aligned} \frac{\cos \mathrm{A}+\cos \mathrm{B}}{\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)} & =\frac{2 \cos \frac{\mathrm{ A}+\mathrm{B}}{2} \cdot \cos \frac{\mathrm{ A}-\mathrm{B}}{2}}{\cos \left(\frac{\mathrm{ A}-\mathrm{B}}{2}\right)} \\ & =2 \cos \left(\frac{\pi-\mathrm{C}}{2}\right) \\ & =2 \cos \left(\frac{\pi}{2}-\frac{\mathrm{C}}{2}\right)=2 \sin \frac{\mathrm{C}}{2} \\ & =2 \sin 30^{{}^{\circ}}=2 \times \frac{1}{2}=1.\end{aligned}
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