Question

How do you find the exact value of `Arctan (-1)`?

Answer 1

`arc tan(-1)=npi+(3pi)/4, n=0, +-1, +-2, +-3...` n=0, gives its value as `(3pi)/4 in [0, pi]`. Including n=1, we get two values `(3pi)/4 and (7pi)/4in [0, 2pi]`.

Explanation:

Use `tan (pi-x)=tan x`.
`tan(pi-pi/4)=-tan(pi/4)=-1`

So, a principal value of `arc tan (-1) in [0, pi] is (3pi)/4`.
There is another #in [0, 2pi]. tan(2pi-pi/4)=-tan (pi/4)=-1#

So, there are two, `(3pi)/4 and (7pi)/4, in [0, 2pi]`.

Either can be taken as principal value, for the general value

`arc tan(-1)=npi+(3pi)/4, n=0, +-1, +-2, +-3...` . .