Question

For all 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. $\text{cos}A\cos B\cos C$
• C. $\sin A\sin B\sin C$
• D. $\cos P\cos Q\cos R$

Answer 1

Answer: (A)

0

Answer 2

The determinant can be expanded as

\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| = \left| \begin{matrix} \cos A & \sin A & 0 \\ \cos B & \sin B & 0 \\ \cos C & \sin C & 0 \end{matrix} \right| \times \left| \begin{matrix} \cos P & \sin P & 0 \\ \cos Q & \sin Q & 0 \\ \cos R & \sin R & 0 \end{matrix} \right| = 0$$