Question

Solve: `tan^-1 ("1 - x"/"1 + x") = 1/2 (tan^-1 "x")`, for x > 0.

Answer 1

`tan^-1 ("1 - x"/"1 + x") = 1/2 (tan^-1 "x")`

∴ `2 tan^-1 ("1 - x"/"1 + x") = (tan^-1 "x")`

∴ `tan^-1  [(2 ("1 - x"/"1 + x"))/(1 - ("1 - x"/"1 + x")^2)] = tan^-1 "x"   ....[because 2 tan^-1 "x" = tan^-1  (("2x")/(1- "x"^2))]`

∴ `(2 ("1 - x"/"1 + x")(1 + "x")^2)/((1 + "x")^2 - (1 - "x")^2) = "x"`

∴ `(2 (1 - "x")(1 + "x"))/((1 + "2x" + "x"^2) - (1 - "2x" + "x"^2)) = "x"`

∴ `(2(1 - "x"^2))/(1 + "2x" + "x"^2 - 1 + 2"x" - "x"^2) = "x"`

∴ `(2 - 2"x"^2)/"4x" = "x"`

∴ 2 - 2x² = 4x²

∴ 6x² = 2

∴ x² = `1/3`

∴ x = `1/sqrt3`   .  ....[∵ x > 0]