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

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(B) Expanding the determinant 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(

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

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(B) Expanding the determinant along the first row:$\cos(A+B) [\cos A \cos B - \sin A \sin B] - (-\sin(A+B)) [\sin A \cos B - (-\cos A \sin B)] + \cos(2B) [\sin A \sin A - (-\cos A \cos A)] = 0$Simplify the terms inside the brackets:$\cos(A+B) [\cos(A+B)] + \sin(A+B) [\sin(A+B)] + \cos(2B) [\sin^2 A + \cos^2 A] = 0$Using the identity $\sin^2 	heta + \cos^2 	heta = 1$:$\cos^2(A+B) + \sin^2(A+B) + \cos(2B) = 1 = 0$Since $\cos^2 	heta + \sin^2 	heta = 1$:$1 + \cos(2B) = 0$$\cos(2B) = -1$$2B = (2n+1)\pi$$B = (2n+1) \frac{\pi}{2}, n \in Z$

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 $B$ is

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