If $A$, $B$ and $C$ are the angles of a triangle then the value of the determinant

$\left| {\begin{array}{*{20}{c}}
  { - 1 + \cos B}&{\cos C + \cos B}&{\cos B} \\ 
  {\cos C + \cos A}&{ - 1 + \cos A}&{\cos A} \\ 
  { - 1 + \cos B}&{ - 1 + \cos A}&{ - 1} 
\end{array}} \right|$

$\left| {\,\begin{array}{*{20}{c}}{a - b - c}&{2a}&{2a}\\{2b}&{b - c - a}&{2b}\\{2c}&{2c}&{c - a - b}\end{array}\,} \right| = $

If $a,b,c$ are different and $\left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{{a^3} - 1}\\b&{{b^2}}&{{b^3} - 1}\\c&{{c^2}}&{{c^3} - 1}\end{array}\,} \right| = 0$, then

Using properties of determinants, prove this:

$\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q\end{array}\right|=1$

If the determinant $\left| {\begin{array}{*{20}{c}}{a\, + \,p}&{1\, + \,x}&{u\, + \,f}\\ {b\, + \,q}&{m\, + \,y}&{v\, + \,g}\\{c\, + \,r}&{n\, + \,z}&{w\, + \,h}\end{array}} \right|$ splits into exactly $K$ determinants of order $3$, each element of which contains only one term, then the value of $K$, is

The solutions of the equation $\left|\begin{array}{ccc}1+\sin ^{2} x & \sin ^{2} x & \sin ^{2} x \\ \cos ^{2} x & 1+\cos ^{2} x & \cos ^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1+4 \sin 2 x\end{array}\right|=0,(0< x< \pi), \operatorname{are}$