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}#.