Q216 / 270 · NDA 2019 (Phase 2) · Mathematics

What is $\cos 80^{\circ}+\cos 40^{\circ}-\cos 20^{\circ}$ equal to?

  • A
    2
  • B
    1
  • Correct answer
    C
    0
  • D
    -19

Solution

$\begin{aligned} & \text { } \because \cos C+\cos D \\ & =2 \cos \left(\frac{C+D}{2}\right) \cdot \cos \left(\frac{C-D}{2}\right) \\ & \Rightarrow \cos 80^{\circ}+\cos 40^{\circ} \\ & =2 \cos 60^{\circ} \cdot \cos 20^{\circ} \\ & =2 \cdot \frac{1}{2} \cdot \cos 20^{\circ}=\cos 20^{\circ} \\ & \therefore \quad \cos 80^{\circ}+\cos 40^{\circ}-\cos 20^{\circ}=0\end{aligned}$

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

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