If left|begin{array}{ccc}cos (A+B) & -sin (A+ | vedclass

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(A) Given the determinant equation:$\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

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

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(A) Given the determinant equation:$\left|\begin{array}{ccc}\cos (A+B) & -\sin (A+B) & \cos 2 B \\sin A & \cos A & \sin B \-\cos A & \sin A & \cos B\end{array}ight|=0$Expanding along the first row:$\cos (A+B)[\cos A \cos B - \sin A \sin B] + \sin (A+B)[\sin A \cos B + \cos A \sin B] + \cos 2 B[\sin^2 A + \cos^2 A] = 0$Using trigonometric identities $\cos(A+B) = \cos A \cos B - \sin A \sin B$ and $\sin(A+B) = \sin A \cos B + \cos A \sin B$:$\cos (A+B) \cos (A+B) + \sin (A+B) \sin (A+B) + \cos 2 B(1) = 0$$\cos^2 (A+B) + \sin^2 (A+B) + \cos 2 B = 0$Since $\cos^2 	heta + \sin^2 	heta = 1$, we get:$1 + \cos 2 B = 0$$\cos 2 B = -1$$2 B = (2 n+1) \pi$$B = \frac{(2 n+1) \pi}{2} = (2 n+1) \frac{\pi}{2}$

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

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