For all values of A, B, C and P, Q, R,the val | vedclass

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(A) The given determinant is $\Delta = \left| \begin{array}{ccc} \cos(A-P) & \cos(A-Q) & \cos(A-R) \ \cos(B-P) & \cos(B-Q) & \cos(B-R) \ \cos(C-P) &

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$0$

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(A) The given determinant is $\Delta = \left| \begin{array}{ccc} \cos(A-P) & \cos(A-Q) & \cos(A-R) \ \cos(B-P) & \cos(B-Q) & \cos(B-R) \ \cos(C-P) & \cos(C-Q) & \cos(C-R) \end{array} ight|$.Using the formula $\cos(x-y) = \cos x \cos y + \sin x \sin y$,we can write each element as a sum of two terms.This allows us to express the determinant as the product of two matrices:$\Delta = \left| \begin{array}{ccc} \cos A & \sin A & 0 \ \cos B & \sin B & 0 \ \cos C & \sin C & 0 \end{array} ight| 	imes \left| \begin{array}{ccc} \cos P & \cos Q & \cos R \ \sin P & \sin Q & \sin R \ 0 & 0 & 0 \end{array} ight|$.Since the third column of the first matrix is all zeros and the third row of the second matrix is all zeros,the product of these two matrices is zero.Alternatively,the determinant can be split into $8$ determinants,each of which has at least two identical columns (or rows),making the value of each determinant $0$.Thus,the value of the determinant is $0$.

For all values of $A, B, C$ and $P, Q, R$,the value of $\left| \begin{array}{ccc} \cos(A-P) & \cos(A-Q) & \cos(A-R) \\ \cos(B-P) & \cos(B-Q) & \cos(B-R) \\ \cos(C-P) & \cos(C-Q) & \cos(C-R) \end{array} \right|$ is

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