AP EAMCET · Maths · Determinants
If \(\left|\begin{array}{ccc}\cos (A+B) & -\sin (A+B) & \cos 2 B \\ \sin A & \cos A & \sin B \\ -\cos A & \sin A & \cos B\end{array}\right|=0\)
then \(B\) is equal to
- A \((2 n+1) \frac{\pi}{2}\)
- B \((2 n+1) \pi\)
- C \(n \pi\)
- D \(2 n \pi\)
Answer & Solution
Correct Answer
(A) \((2 n+1) \frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
We have, \(\left|\begin{array}{ccc}\cos (A+B) & -\sin (A+B) & \cos 2 B \\\sin A & \cos A & \sin B \\-\cos A & \sin A & \cos B\end{array}\right|=0\) \(\Rightarrow \quad \cos (A+B)[\cos A \cos B-\sin A \sin B]\) \(+\sin (A+B)[\sin A \cos B+\cos A \sin B]\)…
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