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 the value of \(\mathrm{B}\) is

\(\begin{aligned} & \left|\begin{array}{ccc}\cos (A+B) & -\sin (A+B) & \cos (2 B) \\ \sin A & \cos A & \sin B \\ -\cos A & \sin A & \cos B\end{array}\right|=0 \\ & \therefore \quad \cos (\mathrm{A}+\mathrm{B})[(\cos \mathrm{A} \cos \mathrm{B}-\sin \mathrm{A} \sin \mathrm{B})] \\ & +\sin (A+B)[\sin A \cos B+\sin B \cos A] \\ & +\cos 2 \mathrm{~B}\left[\sin ^2 \mathrm{~A}+\cos ^2 \mathrm{~A}\right]=0 \\ & \therefore \quad \cos (\mathrm{A}+\mathrm{B}) \cdot \cos (\mathrm{A}+\mathrm{B}) \\ & +\sin (A+B) \cdot \sin (A+B)+\cos 2 B=0 \\ & \cos ^2(A+B)+\sin ^2(A+B)+\cos 2 B=0 \\ & 1+\cos 2 B=0 \\ & 2 \cos ^2 \mathrm{~B}=0 \\ & \therefore \quad \cos \mathrm{B}=0 \\ & \therefore \quad B=(2 n+1) \frac{\pi}{2} \text { for }(n \in Z) \\ & \end{aligned}\)