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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Determinants

NDA 2018 · Mathematics · Determinants

If A, B and C are the angles of a triangle and | array ccc 1 & 1 &...

If A,BandCA, B and C are the angles of a triangle and \left|1111+sin⁡A1+sin⁡B1+sin⁡Csin⁡ A+sin⁡2 Asin⁡ B+sin⁡2 Bsin⁡C+sin⁡2C\begin{array}{ccc} 1 & 1 & 1 \\ 1+\sin \mathrm{A} & 1+\sin \mathrm{B} & 1+\sin \mathrm{C} \\ \sin \mathrm{~A}+\sin ^2 \mathrm{~A} & \sin \mathrm{~B}+\sin ^2 \mathrm{~B} & \sin \mathrm{C}+\sin ^2 \mathrm{C} \end{array}\right|=0 then which one of the following is correct?

Options

  1. AThe triangle ABC is isosceles
  2. BThe triangle ABC is equilateral
  3. CThe triangle ABC is scalene
  4. DNo conclusion can be drawn with regard to the nature of the triangle

Correct answer

A. The triangle ABC is isosceles

Step-by-step solution

\left|1111+sin⁡A1+sin⁡B1+sin⁡Csin⁡A+sin⁡2Asin⁡B+sin⁡2Bsin⁡C+sin⁡2C\begin{array}{ccc}1 & 1 & 1 \\ 1+\sin \mathrm{A} & 1+\sin \mathrm{B} & 1+\sin \mathrm{C} \\ \sin \mathrm{A}+\sin ^2 \mathrm{ A} & \sin \mathrm{ B}+\sin ^2 \mathrm{ B} & \sin \mathrm{C}+\sin ^2 \mathrm{C}\end{array}\right|∣1001+sin⁡Asin⁡A−sin⁡Bsin⁡A−sin⁡Csin⁡A+sin⁡2Asin⁡2A−sin⁡2Bsin⁡2A−sin⁡2C∣+0=0{ Applying C2→C1−C2 and  C3→C1−C3 } =(sin⁡A−sin⁡B)×(sin⁡A−sin⁡C)∣1001+sin⁡A11sin⁡A+sin⁡2A(sin⁡A+sin⁡B)(sin⁡A+sin⁡C)∣=0(sin⁡A−sin⁡B)(sin⁡A−sin⁡C)(sin⁡B−sin⁡C)=0∴sin⁡A=sin⁡B or  sin⁡A=sin⁡C or sin⁡B=sin⁡C either A=B or B=C or A=C. \begin{aligned} & \left|\begin{array}{ccc} 1 & 0 & 0 \\ 1+\sin \mathrm{A} & \sin \mathrm{ A}-\sin \mathrm{B} & \sin \mathrm{ A}-\sin \mathrm{C} \\ \sin \mathrm{ A}+\sin ^2 \mathrm{ A} & \sin ^2 \mathrm{ A}-\sin ^2 \mathrm{ B} & \sin ^2 \mathrm{ A}-\sin ^2 \mathrm{C} \end{array}\right|+0=0 \\ & \left\{\text{ Applying } \mathrm{C}_2 \rightarrow \mathrm{C}_1-\mathrm{C}_2 \text{\text{ and } } \mathrm{C}_3 \rightarrow \mathrm{C}_1-\mathrm{C}_3\right. \text{ \} } \\ & =(\sin \mathrm{A}-\sin \mathrm{B}) \times(\sin \mathrm{A}-\sin \mathrm{C}) \\ & \left|\begin{array}{ccc} 1 & 0 & 0 \\ 1+\sin \mathrm{A} & 1 & 1 \\ \sin \mathrm{ A}+\sin ^2 \mathrm{ A} & (\sin \mathrm{ A}+\sin \mathrm{B}) & (\sin \mathrm{A}+\sin \mathrm{C}) \end{array}\right|=0 \\ & (\sin A-\sin B)(\sin A-\sin C)(\sin B-\sin C)=0 \\ & \therefore \sin \mathrm{ A}=\sin \mathrm{B} \text{\text{ or } } \sin \mathrm{A}=\sin \mathrm{C} \text{ or } \sin \mathrm{B}=\sin \mathrm{C} \text{ either } \mathrm{A}= \\ & \mathrm{B} \text{ or } \mathrm{B}=\mathrm{C} \text{ or } \mathrm{A}=\mathrm{C} \text{. } \end{aligned}$
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