Q191 / 270 · NDA 2024 (Phase 1) · Mathematics

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

  • A
    2
  • B
    $\sqrt{2}$
  • Correct answer
    C
    1
  • D
    $\frac{1}{\sqrt{2}}$

Solution

$\begin{aligned} \frac{\cos \mathrm{A}+\cos \mathrm{B}}{\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)} & =\frac{2 \cos \frac{\mathrm{~A}+\mathrm{B}}{2} \cdot \cos \frac{\mathrm{~A}-\mathrm{B}}{2}}{\cos \left(\frac{\mathrm{~A}-\mathrm{B}}{2}\right)} \\ & =2 \cos \left(\frac{\pi-\mathrm{C}}{2}\right) \\ & =2 \cos \left(\frac{\pi}{2}-\frac{\mathrm{C}}{2}\right)=2 \sin \frac{\mathrm{C}}{2} \\ & =2 \sin 30^{\circ}=2 \times \frac{1}{2}=1 .\end{aligned}$

Hence, the correct answer is option: 'C'.

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