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If $P, Q$ and $R$ are angles of $\Delta P Q R$, then the value of $\left|\begin{array}{ccc}-1 & \cos R & \cos Q \\ \cos R & -1 & \cos P \\ \cos Q & \cos P & -1\end{array}\right|$ is equal to
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$$
\begin{aligned}
&\text { Multiplying } C_{1} \text { by } p \text { and then doing }\\
&\begin{aligned}
C_{1} & \rightarrow C_{1}+q C_{2}+r C_{3} \text { , we get } \\
\Delta &=\frac{1}{p}\left[\begin{array}{ccc}
-p & \cos R & \cos Q \\
p \cos R & -1 & \cos P \\
p \cos Q & \cos P & -1
\end{array} \mid\right.\\
&=\frac{1}{p}\left|\begin{array}{ccc}
-p+q \cos R+r \cos Q & \cos R & \cos Q \\
p \cos R-q+r \cos P & -1 & \cos P \\
p \cos Q+q \cos P-r & \cos P & -1
\end{array}\right| \\
=\frac{1}{p}\left|\begin{array}{ccc}
0 & \cos R & \cos Q \\
0 & -1 & \cos P \\
0 & \cos P & -1
\end{array}\right|=0
\end{aligned}
\end{aligned}
$$
Using the prejection formula $p=q \cos R+r \cos Q$
\begin{aligned}
&\text { Multiplying } C_{1} \text { by } p \text { and then doing }\\
&\begin{aligned}
C_{1} & \rightarrow C_{1}+q C_{2}+r C_{3} \text { , we get } \\
\Delta &=\frac{1}{p}\left[\begin{array}{ccc}
-p & \cos R & \cos Q \\
p \cos R & -1 & \cos P \\
p \cos Q & \cos P & -1
\end{array} \mid\right.\\
&=\frac{1}{p}\left|\begin{array}{ccc}
-p+q \cos R+r \cos Q & \cos R & \cos Q \\
p \cos R-q+r \cos P & -1 & \cos P \\
p \cos Q+q \cos P-r & \cos P & -1
\end{array}\right| \\
=\frac{1}{p}\left|\begin{array}{ccc}
0 & \cos R & \cos Q \\
0 & -1 & \cos P \\
0 & \cos P & -1
\end{array}\right|=0
\end{aligned}
\end{aligned}
$$
Using the prejection formula $p=q \cos R+r \cos Q$
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