If A, B and C are angles of a triangle, then the determinant is equal to
(A) 0
(B) -1
(C) 1
(D) None of these
Option (A) 0
Given,
$\Delta=\left|\begin{array}{ccc}-1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -1\end{array}\right|$
On expanding the determinant, we get
$\Delta=-1+2 \cos A \cos B \cos C+\cos ^{2} A+\cos ^{2} B+\cos ^{2} C$
Now, $2 \cos ^{2} A+2 \cos ^{2} B+2 \cos ^{2} C$
$=1+\cos 2 A+1+\cos 2 B+1+\cos 2 C$
$=3+(\cos 2 A+\cos 2 B+\cos 2 C)$
$=3+(\cos 2 A+\cos 2 B)+\cos 2 C$
$=3+2 \cos (A+B) \cos (A-B)+2 \cos ^{2} C-1$
$=2+2 \cos (\pi-C) \cos (A-B)+2 \cos ^{2} C$
$=2-2 \cos C \cos (A-B)+2 \cos ^{2} C$
$=2-2 \cos C[\cos (A-B)-\cos C]$
$=2-2 \cos C[\cos (A-B)-\cos \{\pi-(A+B)\}]$
$=2-2 \cos C[\cos (A-B)+\cos (A+B)]$
$=2-4 \cos A \cos B \cos C$
$\cos ^{2} A+\cos ^{2} B+\cos ^{2} C=1-2 \cos A \cos B \cos C$
Thus, $\Delta=0$
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