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NDA 2014
Exam
NDA
Year
2014
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2014 · Mathematics · Trigonometric Ratios & Identities

If A+B+C=π , then what is (A+B)+ C equal to?

If A+B+C=π\mathrm{A}+\mathrm{B}+\mathrm{C}=\pi, then what is cos⁡(A+B)+cos⁡C\cos (\mathrm{A}+\mathrm{B})+\cos \mathrm{C} equal to?

Options

  1. A0
  2. B2cos⁡C2 \cos \mathrm{C}
  3. Ccos⁡C−sin⁡C\cos C-\sin C
  4. D2sin⁡C2 \sin \mathrm{C}

Correct answer

A. 0

Step-by-step solution

∵A+B+C=πA+B=π−Ccos⁡(A+B)=cos⁡(π−C)⇒cos⁡(A+B)=−cos⁡C\begin{aligned} \because \quad \mathrm{A}+\mathrm{B}+\mathrm{C} & =\pi \\ \mathrm{A}+\mathrm{B} & =\pi-\mathrm{C} \\ \cos (\mathrm{ A}+\mathrm{B}) & =\cos (\pi-\mathrm{C}) \\ \Rightarrow \cos (\mathrm{A}+\mathrm{B}) & =-\cos \mathrm{C} \end{aligned} Now, cos⁡(A+B)+cos⁡C=−cos⁡C+cos⁡C=0\cos (A+B)+\cos C=-\cos C+\cos C=0
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