Show that $\Delta ABC$ is an isosceles triangle, if the determinant
$\Delta = \left| {\begin{array}{cccccccccccccccccccc}1&1&1\\{1 + \cos A}&{1 + \cos B}&{1 + \cos C}\\{{{\cos }^2}A + \cos A}&{{{\cos }^2}B + \cos B}&{{{\cos }^2}C + \cos C}\end{array}} \right| = 0$

We have $\Delta = \left| {\begin{array}{cccccccccccccccccccc}1&1&1\\{1 + \cos A}&{1 + \cos B}&{1 + \cos C}\\{{{\cos }^2}A + \cos A}&{{{\cos }^2}B + \cos B}&{{{\cos }^2}C + \cos C}\end{array}} \right| = 0$

$\Delta = \left| {\begin{array}{cccccccccccccccccccc}0&0&1\\{\cos A - \cos C}&{\cos B - \cos C}&{1 + \cos C}\\{{{\cos }^2}A + \cos A - {{\cos }^2}C - \cos C}&{{{\cos }^2}B + \cos B - {{\cos }^2}C - \cos C{{\cos }^2}C + \cos C}&{}\end{array}} \right| = 0$

$\left[ {\therefore {C_1} \to {C_1} - {C_3}} \right.$ and $\left. {{C_2} \to {C_2} - {C_3}} \right]$

$\Rightarrow$ $(\cos A - \cos C) \cdot (\cos B - \cos C)$

$\left[ {\begin{array}{cccccccccccccccccccc}0&0&1\\1&1&{1 + \cos C}\\{\cos A + \cos C + 1}&{\cos B + \cos C + 1}&{{{\cos }^2}C + \cos C}\end{array}} \right] = 0$

[ taking $(\cos A - \cos C)$ common from ${C_1}$ and $(\cos B - \cos C)$

common from ${C_2}$]
$\Rightarrow$ $(\cos A - \cos C) \cdot (\cos B - \cos C)[(\cos B + \cos C + 1) - (\cos A + \cos C + 1)] = 0$

$\Rightarrow$ $(\cos A - \cos C) \cdot (\cos B - \cos C)(\cos B + \cos C + 1 - \cos A - \cos C - 1) = 0$

$\Rightarrow$ $(\cos A - \cos C) \cdot (\cos B - \cos C)(\cos B - \cos A) = 0$

i.e. $\cos A = \cos C{\mathop{\rm cor}\nolimits} \cos B = \cos C\,{\rm{or}}\,\cos B = \cos A$

$\Rightarrow$ $A = C$ or $B = C$ or $B = A$

Hence, ABC is an isosceles triangle.