Trigonometric Equations Question 11
Question: If $ | ,\begin{matrix} \cos (A+B) & -\sin (A+B) & \cos 2B \\ \sin A & \cos A & \sin B \\ -\cos A & \sin A & \cos B \\ \end{matrix}, |=0 $ , then B =
[EAMCET 2003]
Options:
A) $ (2n+1)\frac{\pi }{2} $
B) $ n\pi $
C) $ (2n+1)\frac{\pi }{2} $
D) $ 2n\pi $
Show Answer
Answer:
Correct Answer: A
Solution:
- On expanding determinant, $ {{\cos }^{2}}(A+B)+{{\sin }^{2}}(A+B)+\cos 2B=0 $ $ 1+\cos 2B=0 $ or $ \cos 2B=\cos \pi $ or $ 2B=2n\pi +\pi $ or $ B=(2n+1)\frac{\pi }{2},n\in Z. $
BETA
