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Question
In ΔABC, prove that `("a"^2sin("B" - "C"))/(sin"A") + ("b"^2sin("C" - "A"))/(sin"B") + ("c"^2sin("A" - "B"))/(sin"C")` = 0
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Solution
In ∆ABC by sine rule, we have
`(sin"A")/"a" = (sin"B")/"b" = (sin"C")/"c"` = k
∴ sin A = ka, sin B = kb, sin C = kc
Consider a2sin(B − C) = a2(sin B cos C − cos B sin C)
= a2(kb cos C − kc cos B)
= ka(ab cos C − ac cos B)
= `"ak"["ab"(("a"^2 + "b"^2 - "c"^2)/(2"ab")) - "ac"(("a"^2 + "c"^2 - "b"^2)/(2"ac"))]` .......[By consine rule]
= `"ak"[("a"^2 + "b"^2 + "c"^2)/2 - (("a"^2 + "c"^2 - "b"^2)/2)]`
= `"k"/2 ("a")("a"^2 + "b"^2 - "c"^2 - "a"^2 - "c"^2 + "b"^2)`
= `"k"/2 "a"(2"b"^2 - 2"c"^2)`
= ka(b2 − c2)
Similarly, we can prove that
b2sin(C − A) = kb(c2 − a2) and c2sin(A − B)
= kc(a2 − b2)
∴ `("a"^2sin("B" - "C"))/(sin"A") + ("b"^2sin("C" - "A"))/(sin"B") + ("c"^2sin("A" - "B"))/(sin"C")`
= `("ka"("b"^2 - "c"^2))/(sin"A") + ("kb"("c"^2 - "a"^2))/(sin"B") + ("kc"("a"^2 - "b"^2))/(sin"C")`
= `("ka"("b"^2 - "c"^2))/"ka" + ("kb"("c"^2 - "a"^2))/"kb" + ("kc"("a"^2 - "b"^2))/"kc"`
= (b2 − c2 + c2 − a2 + a2 − b2)
= 0
∴ `("a"^2sin("B" - "C"))/(sin"A") + ("b"^2sin("C" - "A"))/(sin"B") + ("c"^2sin("A" - "B"))/(sin"C")` = 0
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