Show that ΔABC is an isosceles triangle, if the determinant

Engineering

Mathematics

Properties of Determinant

Trigonometric Identities in a Triangle

cubic and higher degree equation

Question

Show that ΔABC is an isosceles triangle, if the determinant $∣ \begin{array}{ccc} 1 & 1 & 1 \\ 1 + \cos A & 1 + \cos B & 1 + \cos C \\ \cos^{2} A + \cos A & \cos^{2} B + \cos B & \cos^{2} C + \cos C \end{array} ∣ = 0$

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$∣ \begin{array}{ccc} 1 & 1 & 1 \\ 1 + \cos A & 1 + \cos B & 1 + \cos C \\ \cos^{2} A + \cos A & \cos^{2} B + \cos B & \cos^{2} C + \cos C \end{array} ∣ = 0$

Applying C₁ → C₁ C₃, C₂ → C₂ – C₃, we get,

∣ \begin{array}{ccc} 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} ∣ = 0

∣ \begin{array}{ccc} 0 & 0 & 1 \\ \cos A - \cos C & \cos B - \cos C & 1 + \cos C \\ (\cos A + \cos C + 1) (\cos A - \cos C) & (\cos B + \cos C + 1) (\cos B - \cos C) & \cos^{2} C + \cos C \end{array} ∣ = 0

Taking (cosA – cosC) common from C₁ and (cosB – cosC) common from C₂, we get

(\cos A - \cos C) (\cos B - \cos C) ∣ \begin{array}{ccc} 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} ∣ = 0

Expanding along R₁,

(cos⁡A – cos⁡C)(cos⁡B – cos⁡C)[(cos⁡B + cos⁡C + 1) – (cos⁡A + cos⁡C + 1)] = 0

(cos⁡A – cos⁡C)(cos⁡B – cos⁡C)(cos⁡B – cos⁡A) = 0

cos⁡A = cos⁡C or cos⁡B = cos⁡C or cos⁡B = cos⁡A

A = C or B = C or B = A

Hence, ABC is an isosceles triangle.