Question

In a triangle ABC, if `|(1, 1, 1),(1 + sin"A", 1 + sin"B", 1 + sin"C"),(sin"A" + sin^2"A", sin"B" + sin^2"B", sin"C" + sin^2"C")|` = 0, then prove that ∆ABC is an isoceles triangle.

Answer 1

Let ∆ = `|(1, 1, 1),(1 + sin"A", 1 + sin"B", 1 + sin"C"),(sin"A" + sin^2"A", sin"B" + sin^2"B", sin"C" + sin^2"C")|`

= `|(1, 1, 1),(1 + sin"A", 1 + sin"B", 1 + sin"C"),(-cos^2"A", -cos^2"B", -cos^2"C")|`   R₃ → R₃ – R₂

= `|(1, 0, 0),(1 + sin"A", sin"B" - sin"A", sin"C" - sin"B"),(-cos^2"A", cos^2"A" - cos^2"B", cos^2"B" - cos^2"C")|` .......(C₃ → C₃ – C₂ and C₂ → C₂ – C₁)

Expanding along R₁, we get

∆ = (sinB – sinA)(sin²C – sin²B) – (sinC – sin B)(sin²B – sin²A)

= (sinB – sinA)(sinC – sinB)(sinC – sin A)

= 0

⇒ Either sinB – sinA = 0 or sinC – sinB or sinC – sinA = 0

⇒ A = B or B = C or C = A

i.e. triangle ABC is isoceles.