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प्रश्न
In ∆ABC, prove that `(cos^2"A" - cos^2"B")/("a" + "b") + (cos^2"B" - cos^2"C")/("b" + "c") + (cos^2"C" - cos^2"A")/("c" + "a")` = 0
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उत्तर
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
L.H.S. = `(cos^2"A" - cos^2"B")/("a" + "b") + (cos^2"B" - cos^2"C")/("b" + "c") + (cos^2"C" - cos^2"A")/("c" + "a")`
= `((1 - sin^2"A") - (1 - sin^2"B"))/("a" + "b") + ((1 - sin^2"B") - (1 - sin^2"C"))/("b" + "c") + ((1 - sin^2"C") - (1 - sin^2"A"))/("c" + "a")`
= `(sin^2"B" - sin^2"A")/("a" + "b") + (sin^2"C" - sin^2"B")/("b" + "c") + (sin^2"A" - sin^2"C")/("c" + "a")`
= `("k"^2"b"^2 - "k"^2"a"^2)/("a" + "b") + ("k"^2"c"^2 - ""^2"b"^2)/("b" + "c") + ("k"^2"a"^2 - "k"^2"c"^2)/("c" + "a")`
= `("k"^2("b" - "a")("b" + "a"))/("a" + "b") + ("k"^2("c" - "b")("c" + "b"))/("b" + "c") + ("k"^2("a" - "c")("a" + "c"))/("c" + "a")`
= k2(b − a + c − b + a − c)
= 0
= R.H.S.
∴ `(cos^2"A" - cos^2"B")/("a" + "b") + (cos^2"B" - cos^2"C")/("b" + "c") + (cos^2"C" - cos^2"A")/("c" + "a")` = 0
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