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

Δ = `[(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")]` = 0

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#### Solution

We have, Δ = `[(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")]` = 0

[Applying C_{1} → C_{1} – C_{2 }and C_{2} → C_{2} – C_{3}]

⇒ `[(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")]` = 0

[Taking (cos A – cos C) common from C_{1} and (cos B – cos C) common from C_{2}]

⇒ `(cos "A" - cos "C") (cos "B" - cos "C") xx [(0, 0, 1),(1, 1, 1 + cos"C"),(cos"A" + cos"C" + 1, cos"B" + cos"C" + 1, cos^2"C" + cos"C")]` = 0

[Applying C_{1} → C_{1} – C_{2}]

⇒ `(cos "A" - cos "C") (cos "B" - cos "C") xx [(0, 0, 1),(0, 1, 1 + cos"C"),(cos"A" - cos"B", cos"B" + cos"C" + 1, cos^2"C" + cos"C")]` = 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, ΔABC is an isosceles triangle.

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