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

Answer 1

`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