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

A) 0
B) $\cos{A\cos{B\cos C}}$
C) $\sin{A\sin{B\sin C}}$
D) $\cos{P\cos{Q\cos R}}$
Explanation

(a)
$\left| \begin{matrix}
\cos{(A - P)} & \cos{(A - Q)} & \cos{(A - R)} \\
\cos{(B - P)} & \cos{(B - Q)} & \cos{(B - R)} \\
\cos{(C - P)} & \cos{(C - Q)} & \cos{(C - R)}
\end{matrix} \right|$
$= \left| \begin{matrix}
\cos{A\cos{P + \sin{A\sin P}}} & \cos{A\cos{Q + \sin{A\sin Q}}} & \cos{A\cos{R + \sin{A\sin R}}} \\
\cos{B\cos{P + \sin{B\sin P}}} & \cos{B\cos Q} + \sin{B\sin Q} & \cos{B\cos{R + \sin{B\sin R}}} \\
\cos{C\cos{P + \sin{C\sin P}}} & \cos{C\cos Q} + \sin{C\sin Q} & \cos{C\cos{R + \sin{C\sin R}}}
\end{matrix} \right|$
The determinants can be rewritten as 8 determinants and the value of each of these 8 determinants is zero.
$ie,\cos{P\cos{Q\cos{R\left| \begin{matrix}
\cos A & \cos A & \cos A \\
\cos B & \cos B & \cos B \\
\cos C & \cos C & \cos C
\end{matrix} \right| = 0}}}$
Similarly, other determinants can be shown zero.

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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