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