Q161 / 270 · NDA 2021 (Phase 2) · Mathematics

Let $A B C$ be a triangle. If $\cos 2 A+\cos 2 B+\cos 2 C=-1$ then which one of the following is correct?

  • A
    $\sin A \sin B \sin C=0$
  • B
    $\sin A \sin B \cos C=0$
  • C
    $\cos A \sin B \sin C=0$
  • Correct answer
    D
    $\cos A \cos B \cos C=0$

Solution

$\begin{aligned} & \therefore A+B+C=\pi \\ & \Rightarrow C=\pi-(A+B) \\ & \text {Now, } \cos 2 A+\cos 2 B+\cos 2 C=-1 \\ & \Rightarrow 2 \cos (A+B) \cdot \cos (A-B)+2 \cos ^2 C-1=-1 \\ & \Rightarrow 2 \cos (A+B) \cdot \cos (A-B)+2 \cos ^2(\pi-(A+B))=0 \\ & \Rightarrow 2 \cos (A+B)[\cos (A-B)+\cos (A+B)]=0 \\ & \Rightarrow 2 \cos (\pi-C)[2 \cos A \cdot \cos B]=0 \\ & \Rightarrow-4 \cos C \cdot \cos A \cdot \cos B=0 \\ & \Rightarrow \cos A \cdot \cos B \cdot \cos C=0\end{aligned}$

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

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