If $A,B$and $C$ are angles of a triangle, then the determinant $\left| {\begin{array}{cccccccccccccccccccc}{ - 1}&{\cos C}&{\cos B}\\{\cos C}&{ - 1}&{\cos A}\\{\cos B}&{\cos A}&{ - 1}\end{array}} \right|$ is equal to

We have, $\left| {\begin{array}{cccccccccccccccccccc}{ - 1}&{\cos C}&{\cos B}\\{\cos C}&{ - 1}&{\cos A}\\{\cos B}&{\cos A}&{ - 1}\end{array}} \right|$

Applying ${C_1} \to a{C_1} + b{C_2} + c{C_3}$

$\left| {\begin{array}{cccccccccccccccccccc}{ - a + b\cos C + c\cos B}&{\cos C}&{\cos B}\\{a\cos C - b + c\cos A}&{ - 1}&{\cos A}\\{a\cos B + b\cos A - c}&{\cos A}&{ - 1}\end{array}} \right|$

Also, by projection rule in a triangle,

we know that $a = b\cos C + c\cos B,b = c\cos A + a\cos C$ and $c = a\cos B + b\cos A$

Using above equation in column first,

we get $\left| {\begin{array}{cccccccccccccccccccc}{ - a + a}&{\cos C}&{\cos B}\\{b - b}&{ - 1}&{\cos A}\\{c - c}&{\cos A}&{ - 1}\end{array}} \right|$

$= \left| {\begin{array}{cccccccccccccccccccc}0&{\cos C}&{\cos B}\\0&{ - 1}&{\cos A}\\0&{\cos A}&{ - 1}\end{array}} \right| = 0$

[since, determinant having all elements of any column or row gives value of determinant as zero]