If `tan^(-1)y=4tan^(-1)x(|x|

Explanation

We have `tan^(-1) y = 4 tan^(-1) x`
`rArr tan^(-1) y = tan^(-1) (2x)/(1 -x^(2))` (as `|x| lt1`)
`= tan^(-1). ((4 x)/(1 -x^(2)))/(1 - (4x^(2))/((1 - x^(2))^(2)))`
`= tan^(-1). (4 x (1 - x^(2)))/(x^(4) - 6x^(2) + 1) ("as" |(2x)/(1 -x^(2))| lt 1)`
or `y = (4x (1 - x^(2)))/(x^(4) - 6x^(2) + 1)`
If `x = tan. (pi)/(8)`
`rArr tan^(-1) y = 4 tan^(-1) x = (pi)/(2)`
`rArr y rarr oo`
`rArr x^(4) 6x^(2) + 1 = 0`
Hence, `tan.(pi)/(8)` is a root of `x^(4) - 6x^(2) + 1 = 0`