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प्रश्न
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.
उत्तर
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.
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