Matrices and Determinants 2 Question 35
39. For all values of $A, B, C$ and $P, Q, R$, show that
$(1994,4 M)$
$$ \begin{array}{cccc} \cos (A-P) & \cos (A-Q) & \cos (A-R) & \\ \cos (B-P) & \cos (B-Q) & \cos (B-R) & =0 \\ \cos (C-P) & \cos (C-Q) & \cos (C-R) & \end{array} $$
Show Answer
Solution:
- Let $\Delta=\left|\begin{array}{lll}\cos (A-P) & \cos (A-Q) & \cos (A-R) \ \cos (B-P) & \cos (B-Q) & \cos (B-R) \ \cos (C-P) & \cos (C-Q) & \cos (C-R)\end{array}\right|$
$\Rightarrow \quad \Delta=\mid \begin{array}{ll}\cos A \cos P+\sin A \sin P & \cos (A-Q) \ \cos B \cos P+\sin B \sin P & \cos (B-Q) \ \cos C \cos P+\sin C \sin P & \cos (C-Q)\end{array}$
$$ \cos (A-R) $$
$$ \begin{aligned} & \cos (A-R) \\ & \cos (B-R) \\ & \cos (C-R) \end{aligned} $$
$$ \sin A \sin P \quad \cos (A-Q) \quad \cos (A-R) $$
$+\sin B \sin P \quad \cos (B-Q) \quad \cos (B-R)$ $\sin C \sin P \quad \cos (C-Q) \quad \cos (C-R)$
$\Rightarrow \quad \Delta=\cos P \begin{array}{cccc}\cos A & \cos (A-Q) & \cos (A-R) \ \cos B & \cos (B-Q) & \cos (B-R)\end{array}$ $\begin{array}{lll}\cos C & \cos (C-Q) & \cos (C-R)\end{array}$
$$ \begin{array}{llll} & \sin A & \cos (A-Q) & \cos (A-R) \\ +\sin P & \sin B & \cos (B-Q) & \cos (B-R) \\ & \sin C & \cos (C-Q) & \cos (C-R) \end{array} $$
Applying $C _2 \rightarrow C _2-C _1 \cos Q, C _3 \rightarrow C _3-C _1 \cos R$ in first determinant and $C _2 \rightarrow C _2-C _1 \sin Q$ and in second determinant
$\Rightarrow \Delta=\cos P\left|\begin{array}{lll}\cos A & \sin A \sin Q & \sin A \sin R \ \cos B & \sin B \sin Q & \sin B \sin R \ \cos C & \sin C \sin Q & \sin C \sin R\end{array}\right|$ $\begin{array}{llll} & \sin A & \cos A \cos Q & \cos A \cos R \ +\sin P & \sin B & \cos B \cos Q & \cos B \cos R \ \sin C & \cos C \cos Q & \cos C \cos R\end{array}$
$\begin{array}{llll}\sin A & \cos A & \cos A\end{array}$ $\begin{array}{llll}+\sin P \cos Q \cos R & \sin B & \cos B & \cos B\end{array}$ $\begin{array}{lll}\sin C & \cos C & \cos C\end{array}$
$$ \Delta=0+0=0 $$
$$ n ! \quad(n+1) ! \quad(n+2) ! $$
