Question

Find the domain of the following functions:

`1/2 tan^-1 (1 - x^2) - pi/4`

Answer 1

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

So `1/2 tan^-1 (1 - x^2) = tan^-1 [(1 - (1 - x^2))/(1 + (x - x^2))]`

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

So `1/2 tan^-1(1 - x^2) - pi/4 = tan^-1  x^2/(2 - x^2) - tan^-1 1`

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

= `tan^-1(x^2 - 1)`

= y (say)

The domain of y is `(-oo, oo) {x | x ∈ -1}` and range is `[-1, oo){y | y ≥ -1}`

The domain for tan^–1(x² – 1) is (2n + 1)π. Since tan x is an odd function.