Question

If $\left| \begin{matrix} \cos(A + B) & - \sin(A + B) & \cos 2B \ \sin A & \cos A & \sin B \

• \cos A & \sin A & \cos B \end{matrix} \right| = 0,$then$B =$

• A. $(2n + 1)\frac{\pi}{2}$
• B. $n\pi$
• C. $(2n + 1)\pi$
• D. $2n\pi$

Answer 1

Answer: (A)

$(2n + 1)\frac{\pi}{2}$

Answer 2

On expanding the determinant

$\mathbf{\cos}^{\mathbf{2}}\mathbf{(}\mathbf{A + B) +}\mathbf{\sin}^{\mathbf{2}}\mathbf{(}\mathbf{A + B) +}\mathbf{\cos}\mathbf{2}\mathbf{B = 0}\mathbf{1 +}\mathbf{\cos}\mathbf{2}\mathbf{B = 0}$ or

$\cos 2B = \cos\pi$ or $2B = 2n\pi + \pi$ or $B = (2n + 1)\frac{\pi}{2}.$