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प्रश्न
In a ∆ABC, prove the following, `("a"sin("B" - "C"))/("b"^2 - "c"^2) = ("b"sin("C" - "A"))/("c"^2 - "a"^2) = ("c"sin("A" - "B"))/("a"^2 - "b"^2)`
उत्तर
We know `"a"/sin"A" = "b"/sin"B" = "c"/sin"C"` = 2R
`"a"/sin"A"` = 2R ⇒ a = 2R sin A
`"b"/sin"B"` = 2R ⇒ b = 2R sin B
`"c"/sin"C"` = 2R ⇒ c = 2R sin C
Also sin(A + B) . sin(A – B) = sin2A – sin2B
`("a"sin("B" - "C"))/("b"^2 - "c"^2) = (2"R" sin"A" sin("B" - "C"))/((2"R" sin"B")^2 - (2"R" sin"C")^2`
= `(2"R" sin"A" sin("B" - "C"))/(4"R"^2 sin^2"B" - 4"R"^2 sin^2"C")`
= `(2"R" sin"A" sin("B" - "C"))/(4"R"^2 (sin^2"B" - sin^2"C"))`
= `(2"R" sin"A" sin("B" - "C"))/(4"R"^2 sin("B" + "C") sin("B" - "C"))`
= `sin"A"/(2"R" sin("B" + "C"))`
= `sin"A"/(2"R"sin(180^circ - "A"))`
= `sin"A"/(2"R" sin "A")`
`("a"sin("B" - "C"))/("b"^2 - "c"^2) = 1/(2"R")` ......(1)
`("b"sin("C" - "A"))/("c"^2 - "a"^2) = (2"R" sin"B" sin("C" - "A"))/((2"R" sin"C")^2 - (2"R"sin"A")^2`
= `(2"R" sin"B" sin("C" - "A"))/(4"R"^2 sin^2"C" - 4"R"^2 sin^2"A")`
= `(2"R" sin"B" * sin("C" - "A"))/(4"R"^2 (sin^2"C" - sin^2"A"))`
= `(sin"B" * sin("C" - "A"))/(2"R" sin("C" + "A") sin("C" - "A"))`
= `(sin"B" * sin("C" - "A"))/(2"R" * sin("C" + "A") * sin("C" - "A"))`
= `sin "B"/(2"R"sin("C" + "A"))`
= `sin "B"/(2"R" sin(180^circ - "B"))`
= `sin"B"/(2"R" sin "B")`
`("b"sin("C" - "A"))/("c"^2 - "a"^2) = 1/(2"R")` ......(2)
`("c"sin("A" - "B"))/("a"^2 - "b"^2) = (2"R" sin"C" sin("A" - "B"))/((2"R" sin"A")^2 - (2"R"sin "B")^2)`
= `(2"R" sin"C" sin("A" - "B"))/(4"R"^2 sin^2"A" - 4"R"^2 sin^2"B")`
= `(2"" sin"C" sin("A" - "B"))/(4"R"^2 (sin^2"A" - sin^2"B"))`
= `(2"R" sin"C" sin("A" - "B"))/(4"R"^2 sin("A" + "B") sin("A" - "B"))`
= `sin"C"/(2"R"sin("A" +"B"))`
= `sin"C"/(2"R"sin(180^circ - "C"))`
= `sin"C"/(2"R" sin"C")`
`("c"sin("A" - "B"))/("a"^2 - "b"^2) = 1/(2"R")` ......(3)
From equations (1), (2) and (3)
`("a"sin("B" - "C"))/("b"^2 - "c"^2) = ("b"sin("C" - "A"))/("c"^2 - "a"^2) = ("c"sin("A" - "B"))/("a"^2 - "b"^2)`
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