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Show that the Δ ABC is an isosceles triangle if the determinant

\Delta=\left[\begin{array}{ccc} 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

 

Answers (1)

\\\begin{aligned} &\Delta=\left|\begin{array}{ccc} 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\\ &\text { Apply, } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}\\ &=\left|\begin{array}{ccc} 1 & 1-1 & 1 \\ 1+\cos A & 1+\cos B-1-\cos A & 1+\cos C \\ \cos ^{2} A+\cos A & \cos ^{2} B+\cos B-\cos ^{2} A-\cos A & \cos ^{2} C+\cos C \end{array}\right|=0\\ &=\left|\begin{array}{ccc} 1 & 0 & 1 \\ 1+\cos A & \cos B-\cos A & 1+\cos C \\ \cos ^{2} A+\cos A & \cos ^{2} B+\cos B-\cos ^{2} A-\cos A & \cos ^{2} C+\cos C \end{array}\right|=0\\ &=\left|\begin{array}{ccc} 1 & 0 & 1 \\ 1+\cos A & \cos B-\cos A & 1+\cos C \\ \cos ^{2} A+\cos A & (\cos B-\cos A)(\cos B+\cos A)+(\cos B-\cos A) & \cos ^{2} C+\cos C \end{array}\right|=0 \end{aligned}

\\\begin{aligned} &\begin{array}{|ccc|} 1 & 0 & 1 \\ 1+\cos A & \cos B-\cos A & 1+\cos C \\ \cos ^{2} A+\cos A & (\cos B-\cos A)[(\cos B+\cos A)+1] & \cos ^{2} C+\cos C \end{array} \mid=0\\ &\text { Take, cosB-cosA common from column } 2\\ &\Rightarrow \cos B-\cos A\left|\begin{array}{ccc} 1 & 0 & 1 \\ 1+\cos A & 1 & 1+\cos C \\ \cos ^{2} A+\cos A & (\cos B+\cos A)+1 & \cos ^{2} C+\cos C \end{array}\right|=0\\ &\text { Apply } \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}\\ &\cos B-\cos A\left|\begin{array}{ccc} 1 & 0 & 1-1 \\ 1+\cos A & 1 & 1+\cos C-1-\cos A \\ \cos ^{2} A+\cos A & \cos B+\cos A+1 & \cos ^{2} C+\cos C-\cos ^{2} A-\cos A \end{array}\right|=0 \end{aligned}

\\\begin{aligned} &\Rightarrow \cos B-\cos A\left|\begin{array}{ccc} 1 & 0 & 0 \\ 1+\cos A & 1 & \cos C-\cos A \\ \cos ^{2} A+\cos A & \cos B+\cos A+1 & \cos ^{2} C-\cos ^{2} A+\cos C-\cos A \end{array}\right|=0\\ &\Rightarrow(\cos B-\cos A)\left|\begin{array}{ccc} 1 & 0 & 0 \\ 1+\cos A & 1 & (\cos C-\cos A) \\ \cos ^{2} A+\cos A & \cos B+\cos A+1 & (\cos C-\cos A)(\cos C+\cos A+1) \end{array}\right|=0\\ &\text { Take cosC-cosA common from Column } 3\\ &\Rightarrow(\cos B-\cos A)(\cos C-\cos A)\left|\begin{array}{ccc} 1 & 0 & 0 \\ 1+\cos A & 1 & 1 \\ \cos ^{2} A+\cos A & \cos B+\cos A+1 & \cos C+\cos A+1 \end{array}\right|=0 \end{aligned}

 

Expand along Row 1

\begin{align*} &\Rightarrow (\cos B - \cos A)(\cos C - \cos A)[(1) \{ \cos C + \cos A + 1 - (\cos B + \cos A + 1) \}] = 0 \\ &\Rightarrow (\cos B - \cos A)(\cos C - \cos A)(\cos C + \cos A + 1 - \cos B - \cos A - 1) = 0 \\ &\Rightarrow (\cos B - \cos A)(\cos C - \cos A)(\cos C - \cos B) = 0 \\ &\Rightarrow \cos B - \cos A = 0 \quad \text{or} \quad \cos C - \cos A = 0 \quad \text{or} \quad \cos C - \cos B = 0 \\ &\Rightarrow \cos B = \cos A \quad \text{or} \quad \cos C = \cos A \quad \text{or} \quad \cos C = \cos B \\ &\Rightarrow B = A \quad \text{or} \quad C = A \quad \text{or} \quad C = B \\ \end{align*}

Hence, ΔABC is an isosceles triangle.

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