Solve the equations.`tan^(-1)((1-x)/(1+x))=1/2tan^(-1)x ,(x >0)`

Answer

`x = (1)/(sqrt3)`

Explanation

`tan^(-1) ((1 -x)/(1 + x)) = (1)/(2) tan^(-1) x, x gt 0`
If `x in (0,1)`
`tan^(-1) ((1 -x)/(1 + x)) = tan^(-1) 1 - tan^(-1) x`
Therefore, given equation reduces to
`tan^(-1) 1 - tan^(-1)x = (1)/(2) tan^(-1) x`
`rArr (pi)/(4) = (3)/(2) tan^(-1)x`
`rArr (pi)/(6) = tan^(-1) x`
`:. x = (1)/(sqrt3)`
For `x in (1, oo)`
`tan^(-1) ((1 - x)/(1 + x)) = pi + tan^(-1) 1 - tan^(-1) x = (5pi)/(4) - tan^(-1) x`
Therefore, given equation reduces to
`(5 pi)/(4) - tan^(-1) x = (1)/(2) tan^(-1) x`
or `(5pi)/(6) = tan^(-1) x`, which is not possible,
Hence, only one solution is `x = (1)/(sqrt3)`
Alternate Solution
`tan^(-1) ((1-x)/(1 + x)) = (1)/(2) tan^(-1) x`
`rArr 2 tan^(-1) ((1 - x)/(1 + x)) = tan^(-1) x`
`rArr tan^(-1).(2((1- x)/(1 + x)))/(1 - ((1 - x)/(1 + x))^(2)) = tan^(-1) x`
`rArr tan^(-1).(2(1 -x^(2)))/(4x) = tan^(-1) x`
`rArr 1 -x^(2) = 2x^(2)`
`rArr 3x^(2) =1`
`:. x = +- (1)/(sqrt3)`
But for `x = -(1)/(sqrt3)`, L.H.S. of (1) is `gt 0`, and R.H.S. is `lt 0`