Show that is an isosceles triangle, if the determinant = 1&1&1 1+… - Vidyadip

Question

Show that $\triangle\text{ABC}$ is an isosceles triangle, if the determinant $\triangle=\begin{vmatrix}1&1&1\\1+\cos\text{A}&1+\cos\text{B}&1+\cos\text{C}\\\cos^2\text{A}+\cos\text{A}&\cos^2\text{B}+\cos\text{B}&\cos^2\text{C}+\cos\text{C}\end{vmatrix}=0.$

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Answer

We have, $\triangle=\begin{vmatrix}1&1&1\\1+\cos\text{A}&1+\cos\text{B}&1+\cos\text{C}\\\cos^2\text{A}+\cos\text{A}&\cos^2\text{B}+\cos\text{B}&\cos^2\text{C}+\cos\text{C}\end{vmatrix}=0$
$ \triangle=\begin{vmatrix}0&0&0\\\cos\text{A}-\cos\text{C}&\cos\text{B}-\cos\text{C}&1+\cos\text{C}\\\cos^2\text{A}+\cos\text{A}-\cos^2\text{C}-\cos\text{C}&\cos^2\text{B}+\cos\text{B}-\cos^2\text{C}-\cos\text{C}&\cos^2\text{C}+\cos\text{C}\end{vmatrix}=0.$
$[\text{C}_1\rightarrow\text{C}_1-\text{C}_3\text{ and C}_2\rightarrow\text{C}_2+\text{C}_3]$
$=\begin{vmatrix}0&0&1\\1 & 1&1+\cos\text{C}\\\cos\text{A}+\cos\text{C}+1&\cos\text{B}+\cos\text{C}+1&\cos^2\text{C}+\cos\text{C}\end{vmatrix}=0$
$\big[\text{Taking }(\cos\text{A}\cos\text{C})\text{ common from C}_1\text{ and }(\cos\text{B}-\cos\text{C})\text{ common from C}_2\big]$
$\Rightarrow\ (\cos\text{A}-\cos\text{C}).(\cos\text{B}-\cos\text{C})\big[(\cos\text{B}+\cos\text{C}+1)-(\cos\text{A}+\cos\text{C}+1)\big]=0$
$\Rightarrow\ (\cos\text{A}-\cos\text{C}).(\cos\text{B}-\cos\text{C})(\cos\text{B}+\cos\text{C}+1-\cos\text{A}-\cos\text{C}-1)=0$
$\Rightarrow\ (\cos\text{A}-\cos\text{C}).(\cos\text{B}-\cos\text{C})(\cos\text{B}-\cos\text{A})=0$
$\text{i.e., }\cos\text{A}=\cos\text{C or }\cos\text{B}=\cos\text{C or}\cos\text{B}=\cos\text{A}$
$\Rightarrow\ \text{A = C or B = C or B = A}$
Hence, ABC is an isosceles triangle.

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