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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2024 · Mathematics · Trigonometric Ratios & Identities

If A B C is a triangle, then what is the value of the determinant |...

If ABCA B C is a triangle, then what is the value of the determinant \left|cos⁡Csin⁡ B0tan⁡ A0sin⁡ B0tan⁡( B+C)cos⁡C\begin{array}{ccc}\cos C & \sin \mathrm{~B} & 0 \\ \tan \mathrm{~A} & 0 & \sin \mathrm{~B} \\ 0 & \tan (\mathrm{~B}+\mathrm{C}) & \cos \mathrm{C}\end{array}\right|$ ?

Options

  1. A-1
  2. B0
  3. C1
  4. D3

Correct answer

B. 0

Step-by-step solution

\left|cos⁡Csin⁡B0tan⁡A0sin⁡B0tan⁡(B+C)cos⁡C\begin{array}{ccc}\cos C & \sin B & 0 \\ \tan A & 0 & \sin B \\ 0 & \tan (B+C) & \cos C\end{array}\right|=cos⁡C[−sin⁡B⋅tan⁡(B+C)=−sin⁡B⋅cos⁡C[sin⁡(B+C)cos⁡(B+C)+sin⁡Acos⁡A]=−sin⁡B⋅cos⁡C[sin⁡(B+C)cos⁡A+sin⁡Acos⁡(B+C)cos⁡A⋅cos⁡(B+C)]=−sin⁡B⋅cos⁡Csin⁡(A+B+C)cos⁡A⋅cos⁡(B+C)=−sin⁡B⋅cos⁡Csin⁡πcos⁡A⋅cos⁡(B+C)=0.\begin{aligned} & =\cos C[-\sin B \cdot \tan (B+C) \\ & =-\sin B \cdot \cos C\left[\frac{\sin (B+C)}{\cos (B+C)}+\frac{\sin A}{\cos A}\right] \\ & =-\sin B \cdot \cos C\left[\frac{\sin (B+C) \cos A+\sin A \cos (B+C)}{\cos A \cdot \cos (B+C)}\right] \\ & =-\sin B \cdot \cos C \frac{\sin (A+B+C)}{\cos A \cdot \cos (B+C)} \\ & =\frac{-\sin B \cdot \cos C \sin \pi}{\cos A \cdot \cos (B+C)}=0 .\end{aligned}$$
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