Let tan^(-1) (1) = theta. That means that tan(theta) = 1.
tan(theta) = 1
sin (theta)/cos(theta) = 1
sin (theta) = cos (theta)
One of the basic identities of trigonometry is sin^color(red)2 (x) + cos^color(red)2 (x) = 1.
If sin (theta) = cos (theta), then sin^color(red)2 (theta) = cos^color(red)2 (theta).
sin^2 (theta) + cos^2 (theta) =1
sin^2 (theta) + sin^2 (theta) = 1
2sin^2(theta) = 1
sin^2(theta) = 1/2
sin(theta) = +- 1/sqrt 2
Rationalize the denominator :
sin (theta) = +- sqrt2/2
We know that theta = pi/4 is a solution to this, so we have to deduce the other ones from it.
There are 4 basic identities which you can prove using the unit circle :
sin (x) = sin(pi-x)
sin(x) = -sin(x-pi)
sin(x) = -sin(2pi-x)
From these, you can get some more solutions for sin(theta) = +- sqrt2/2 :
theta in {(color(red)1pi)/4, (color(red)3pi)/4, (color(red)5pi)/4, (color(red)7pi)/4, ...}
All the highlighted numbers are color(red)"odd" numbers, so we get the set of solutions to be
theta in {((color(red)(2n-1))pi)/4 , n in ZZ} are all the values of theta for which sin(theta) = +- sqrt2/2 and, consequently, tan (theta) = 1. Finally, we get
tan^(-1) (1) = theta , forall theta in {((color(red)(2n-1))pi)/4 , n in ZZ}.
