If left| {begin{array}{*{20}{c}}{cos (A + B)} | vedclass

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(A) Let the determinant be $\Delta$. Expanding along the first row:$\Delta = \cos(A+B)(\cos A \cos B - \sin A \sin B) + \sin(A+B)(\sin A \cos B + \cos

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

Vedclass · Answer

(A) Let the determinant be $\Delta$. Expanding along the first row:$\Delta = \cos(A+B)(\cos A \cos B - \sin A \sin B) + \sin(A+B)(\sin A \cos B + \cos A \sin B) + \cos 2B(\sin A \sin A - (-\cos A \cos A)) = 0$Using trigonometric identities:$\Delta = \cos(A+B)\cos(A+B) + \sin(A+B)\sin(A+B) + \cos 2B(\sin^2 A + \cos^2 A) = 0$$\Delta = \cos^2(A+B) + \sin^2(A+B) + \cos 2B(1) = 0$$1 + \cos 2B = 0$$\cos 2B = -1$Since $\cos 	heta = -1$ implies $	heta = (2n+1)\pi$,we have $2B = (2n+1)\pi$.Therefore,$B = (2n+1)\frac{\pi}{2}$ for $n \in \mathbb{Z}$.

If $\left| {\begin{array}{*{20}{c}}{\cos (A + B)}&{ - \sin (A + B)}&{\cos 2B}\\{\sin A}&{\cos A}&{\sin B}\\{ - \cos A}&{\sin A}&{\cos B}\end{array}} \right| = 0$,then $B =$

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