The determinant left| begin{array}{ccc} cos(t | vedclass

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(B) Let the determinant be $\Delta$. Expanding along the first row $(R_1)$: $\Delta = \cos(	heta + \phi) [\cos 	heta \cos \phi - \sin 	heta \sin \p

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independent of $	heta$

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(B) Let the determinant be $\Delta$. Expanding along the first row $(R_1)$: $\Delta = \cos(	heta + \phi) [\cos 	heta \cos \phi - \sin 	heta \sin \phi] + \sin(	heta + \phi) [\sin 	heta \cos \phi + \cos 	heta \sin \phi] + \cos 2\phi [\sin^2 	heta + \cos^2 	heta]$ Using the trigonometric identities $\cos(A+B) = \cos A \cos B - \sin A \sin B$ and $\sin(A+B) = \sin A \cos B + \cos A \sin B$: $\Delta = \cos(	heta + \phi) \cos(	heta + \phi) + \sin(	heta + \phi) \sin(	heta + \phi) + \cos 2\phi (1)$ $\Delta = \cos^2(	heta + \phi) + \sin^2(	heta + \phi) + \cos 2\phi$ Since $\cos^2 x + \sin^2 x = 1$: $\Delta = 1 + \cos 2\phi$ This expression depends only on $\phi$ and is independent of $	heta$.

The determinant $\left| \begin{array}{ccc} \cos(\theta + \phi) & -\sin(\theta + \phi) & \cos 2\phi \\ \sin \theta & \cos \theta & \sin \phi \\ -\cos \theta & \sin \theta & \cos \phi \end{array} \right|$ is :

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