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

Correct answer: D. A B C=0 aligned & A+B+C=π & C=π-(A+B) & Now, 2 A+ 2 B+ 2 C=-1 & 2 (A+B) · (A-B)+2 ^2 C-1=-1 & 2 (A+B) · (A-B)+2 ^2(π-(A+B))=0 & 2 (A+B)[ (A-B)+ (A+B)]=0 & 2 (π-C)[2 A · B]=0 & -4 C · A · B=0 & A · B · C=0 aligned

NDA 2021

Exam: NDA
Year: 2021
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

NDA 2021 · Mathematics · Trigonometric Ratios & Identities

Let A B C be a triangle. If 2 A+ 2 B+ 2 C=-1 then which one of the...

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?

Options

A $\sin A \sin B \sin C=0$
B $\sin A \sin B \cos C=0$
C $\cos A \sin B \sin C=0$
D $\cos A \cos B \cos C=0$

Correct answer

D. $\cos A \cos B \cos C=0$

Step-by-step 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}

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