SATHEE: Matrices and Determinants 2 Question 35

Matrices and Determinants 2 Question 35

39. For all values of $A, B, C$ and $P, Q, R$ , show that

$(1994, 4 M)$

$\begin{array}{cccc} \cos (A - P) & \cos (A - Q) & \cos (A - R) \\ \cos (B - P) & \cos (B - Q) & \cos (B - R) & = 0 \\ \cos (C - P) & \cos (C - Q) & \cos (C - R) \end{array}$

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Solution:

Let $Δ = | \begin{array}{lll} \cos (A - P) & \cos (A - Q) & \cos (A - R) \cos (B - P) & \cos (B - Q) & \cos (B - R) \cos (C - P) & \cos (C - Q) & \cos (C - R) \end{array} |$

$\Rightarrow Δ =∣ \begin{array}{ll} \cos A \cos P + \sin A \sin P & \cos (A - Q) \cos B \cos P + \sin B \sin P & \cos (B - Q) \cos C \cos P + \sin C \sin P & \cos (C - Q) \end{array}$

$\cos (A - R)$

$\begin{aligned} \cos (A - R) \\ \cos (B - R) \\ \cos (C - R) \end{aligned}$

$\sin A \sin P \cos (A - Q) \cos (A - R)$

$+ \sin B \sin P \cos (B - Q) \cos (B - R)$ $\sin C \sin P \cos (C - Q) \cos (C - R)$

$\Rightarrow Δ = \cos P \begin{array}{cccc} \cos A & \cos (A - Q) & \cos (A - R) \cos B & \cos (B - Q) & \cos (B - R) \end{array}$ $\begin{array}{lll} \cos C & \cos (C - Q) & \cos (C - R) \end{array}$

$\begin{array}{llll} \sin A & \cos (A - Q) & \cos (A - R) \\ + \sin P & \sin B & \cos (B - Q) & \cos (B - R) \\ \sin C & \cos (C - Q) & \cos (C - R) \end{array}$

Applying $C_{2} \to C_{2} - C_{1} \cos Q, C_{3} \to C_{3} - C_{1} \cos R$ in first determinant and $C_{2} \to C_{2} - C_{1} \sin Q$ and in second determinant

$\Rightarrow Δ = \cos P | \begin{array}{lll} \cos A & \sin A \sin Q & \sin A \sin R \cos B & \sin B \sin Q & \sin B \sin R \cos C & \sin C \sin Q & \sin C \sin R \end{array} |$ $\begin{array}{llll} \sin A & \cos A \cos Q & \cos A \cos R + \sin P & \sin B & \cos B \cos Q & \cos B \cos R \sin C & \cos C \cos Q & \cos C \cos R \end{array}$

$\begin{array}{llll} \sin A & \cos A & \cos A \end{array}$ $\begin{array}{llll} + \sin P \cos Q \cos R & \sin B & \cos B & \cos B \end{array}$ $\begin{array}{lll} \sin C & \cos C & \cos C \end{array}$