If `sin^(-1)(x^2+2x+2)+tan^(-1)(x^2-3x-k^2)>pi/2,` then find the values of `kdot`

Explanation

`sin^(-1) (x^(2) + 2x + 2) + tan^(-1) (x^(2) - 3x - k^(2)) gt (pi)/(2)`
`rArr sin^(-1) ((x + 1)^(2) + 1) + tan^(-1) (x^(2) - 3x - k^(2) gt (pi)/(2)`
Now, `sin^(-1) ((x + 1)^(2) + 1)` is defined if `(x + 1)^(2) = 0 " or " x = -1` Putting `x = -1`, we get
`(pi)/(2) + tan^(-1) (4 - k^(2)) gt (pi)/(2)`
`rArr tan^(-1) (4 - k^(2)) gt 0`
`rArr 4 - k^(2) gt 0`
`rArr -2 lt k lt 2`