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

#### 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

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")`

= k^{2}(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