If A, B and C are angles of a triangle,

Solution:

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, $\Delta=0$