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प्रश्न
In Δ ABC, prove that a2 (cos2 B - cos2 C) + b2 (cos2 C - cos2 A) + c2 (cos2 A - cos2 B) = 0.
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उत्तर
By sine rule,
`"a"/"sin A" = "b"/"sin B" = "c"/"sin C"` = k
∴ a = k sin A, b = k sin B, c = k sin C
LHS = `"a"^2 (cos^2"B" - cos^2"C") + "b"^2(cos^2 "C" - cos^2 "A") + "c"^2 (cos^2"A" - cos^2"B")`
`= "k"^2 sin^2"A" [(1 - sin^2"B") - (1 - sin^2"C")] + "k"^2 sin^2"B" [(1 - sin^2"C") - (1 - sin^2"A")] + "k"^2 sin^2"C" [(1 - sin^2"A") - (1 - sin^2"B")]`
`= "k"^2 sin^2"A" (sin^2"C" - sin^2"B") + "k"^2sin^2"B"(sin^2"A" - sin^2"C") + "k"^2 sin^2"C" (sin^2"B" - sin^2"A")`
`= "k"^2 (sin^2"A" sin^2"C" - sin^2"A" sin^2"B" + sin^2"A" sin^2"B" - sin^2"B" sin^2"C" + sin^2"B" sin^2"C" - sin^2"A" sin^2"C")`
`= "k"^2 (0)`
= 0
= RHS.
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