Using properties of determinants, show that is isosceles if: 1 & … - Vidyadip

Question

Using properties of determinants, show that $\triangle\text{ABC}$ is isosceles if:
$\begin{vmatrix} 1 & 1 & 1 \\ 1 + \cos\text{A} & 1 + \cos\text{B} & 1 + \cos\text{C} \\ \cos^{2}\text{A} + \cos\text{A} & \cos^{2}\text{B}+\cos\text{B} & \cos^{2}\text{C} + \cos\text{C} \end{vmatrix} = 0 $

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Answer

$\begin{vmatrix} 1 & 1 & 1 \\ 1 + \cos\text{A} & 1 + \cos\text{B} & 1 + \cos\text{C} \\ \cos^{2}\text{A} + \cos\text{A} & \cos^{2}\text{B}+\cos\text{B} & \cos^{2}\text{C} + \cos\text{C} \end{vmatrix} = 0 $
$\text{Apply }\text{C}_{2}\rightarrow\text{C}_{2} -\text{C}_{1}, \text{C}_{3}\rightarrow\text{C}_{3} - \text{C}_{1}$
$\Leftrightarrow \begin{vmatrix} 1 & 0 & 0 \\ 1 +\cos\text{A} & \cos\text{B} - \cos\text{A} & \cos\text{C} - \cos\text{A} \\ \cos^{2}\text{A}+\cos\text{A} & (\cos\text{B} - \cos\text{A})(\cos\text{B} + \cos\text{A} + 1) & (\cos\text{C} - \cos\text{A}) (\cos\text{C} + \cos\text{A} + 1) \end{vmatrix} = 0 $
$\text{Taking}(\cos\text{B} - \cos\text{A}), (\cos\text{C} - \cos\text{A}) \text{common from C}_{2} \& \text{ C}_{3}$
$\Leftrightarrow(\cos\text{B} - \cos\text{A})(\cos\text{C} - \cos\text{A}) \begin{vmatrix} 1 & 0 & 0 \\ 1 + \cos\text{A} & 1 & 1 \\ \cos^{2}\text{A} + \cos\text{A} & \cos\text{B} + \cos\text{A} + 1 & \cos\text{C} + \cos\text{A} + 1 \end{vmatrix} = 0 $
$\text{Expand along R}_{1}$
$\Leftrightarrow(\cos\text{B} - \cos\text{A})(\cos\text{C} - \cos\text{A}) (\cos\text{C} - \cos\text{B}) = 0$
$ \begin{matrix} \Leftrightarrow\cos\text{A} = \cos\text{B} & \Leftrightarrow\text{A = B} &\Leftrightarrow\triangle\text{ABC is an isosceles triangle} \\ \text{or} & \text{or} \\ \cos\text{B} = \cos\text{C} & \text{B = C} \\ \text{or} & \text{or} \\ \cos\text{C} = \cos\text{A} & \text{C = A} \end{matrix} $

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