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Question
If x = `"pab"/("a + b")`, then prove that `("x + pa")/("x - pa") + ("x + pb")/("x - pb") = (2("a"^2 - "b"^2))/"ab"`
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Solution
x = `"pab"/("a + b") => "x"/"pa" = "b"/("a + b")`
Applying componendo and dividendo
`("x + pa")/("x - pa") = ("b + a + b")/("b - a - b") = ("2b + a")/"- a"` .............(i)
Again , x = `"pab"/("a + b") => "x"/"pb" = "a"/("a + b")`
Applying componendo and dividendo
`("x + pb")/("x - pb") = ("a + a + b")/("a - a - b") = ("2a + b")/"- b"` .............(ii)
Adding (i) and (ii)
`("x + pa")/("x - pa") + ("x + pb")/("x - pb") = ("2b + a")/"- a" +("2a + b")/"- b" = ("a - 2"b"")/"a" + ("b - 2a")/"b"`
`= (-2"b"^2 + "ab" - 2"a"^2 + "ab")/"ab"`
`= (-2"b"^2 + 2 "ab" - 2"a"^2)/"ab"`
`= (2 ("a"^2 - "b"^2))/"ab"` = 2 = RHS
LHS = RHS
Hence proved.
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