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For all values of $A,B,C$ and $P,Q,R$, the value of $\left| {\,\begin{array}{*{20}{c}}{\cos (A - P)}&{\cos (A - Q)}&{\cos (A - R)}\\{\cos (B - P)}&{\cos (B - Q)}&{\cos (B - R)}\\{\cos (C - P)}&{\cos (C - Q)}&{\cos (C - R)}\end{array}\,} \right|$ is
$0$
$\cos A\cos B\cos C$
$sin$ $A$ $sin$ $B$ $sin$ $C$
$cos$ $P$ $cos$ $Q$ $cos$ $R$
Solution
(a) The determinant can be expanded as
$\left| {\begin{array}{*{20}{c}}{\cos A\cos P + \sin A\sin P}&{\cos A\cos Q + \sin A\sin Q}&{}\\{\cos B\cos P + \sin B\sin P}&{\cos B\cos Q + \sin B\sin Q}&{}\\{\cos C\cos P + \sin C\sin P}&{\cos C\cos Q + \sin C\sin Q}&{}\end{array}} \right.$$\left. {\begin{array}{*{20}{c}}{\cos A\cos R + \sin A\sin R}\\{\cos B\cos R + \sin B\sin R}\\{\cos C\cos R + \sin C\sin R}\end{array}\,} \right|$
This determinant can be written as $8$ determinants and the value of each of these $8$ determinants is zero;
e.g., $\cos P\cos Q\cos R{\rm{ }}\left| {\,\begin{array}{*{20}{c}}{\cos A}&{\cos A}&{\cos A}\\{\cos B}&{\cos B}&{\cos B}\\{\cos C}&{\cos C}&{\cos C}\end{array}\,} \right| = 0$
Similarly other determinants can be shown zero.
