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Question
If `tan^-1 ((x - 1)/(x + 1)) + tan^-1 ((2x - 1)/(2x + 1)) = tan^-1 (23/36)` = then prove that 24x2 – 23x – 12 = 0
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Solution
`tan^-1 ((x - 1)/(x + 1)) + tan^-1 ((2x - 1)/(2x + 1)) = tan^-1 (23/36)`
`\implies tan^-1 {((x - 1)/(x + 1) + (2x - 1)/(2x + 1))/(1 - ((x - 1)/(x + 1))((2x - 1)/(2x + 1)))} = tan^-1 (23/36)`
`\implies tan^-1 ((2x^2 - x - 1 + 2x^2 + x - 1)/(2x^2 + 3x + 1 - 2x^2 + 3x - 1)) = tan^-1 (23/36)` ...`{{:("Using formula:"),(tan^-1"a" + tan^-1"b" = tan^-1(("a" + "b")/(1 - "ab"))):}}`
`\implies tan^-1 ((4x^2 - 2)/(6x)) = 23/36`
∴ `(4x^2 - 2)/(6x) = 23/36`
`\implies` 6(4x2 – 2) = 23x
`\implies` 24x2 – 23x – 12 = 0
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