In ΔABC, prove that `("a"^2sin("B" - "C"))/(sin"A") + ("b"^2sin("C" - "A"))/(sin"B") + ("c"^2sin("A" - "B"))/(sin"C")`

#### 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 a^{2}sin (B − C) = a^{2}(sin B cos C − cos B sin C)

= a^{2}(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(b^{2} − c^{2})

Similarly, we can prove that

b^{2}sin (C − A) = kb(c^{2} − a^{2}) and c^{2}sin(A − B)

= kc(a^{2} − b^{2})

∴ `("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"`

= (b^{2} − c^{2} + c^{2} − a^{2} + a^{2} − b^{2})

= 0

∴ `("a"^2sin("B" - "C"))/(sin"A") + ("b"^2sin("C" - "A"))/(sin"B") + ("c"^2sin("A" - "B"))/(sin"C")` = 0