sin^2x = 1-(1 + y)^2/(4y)= (1-y)^2/(4y)
=> tan^2(x) = (1-y)^2/(1+y)^2
=> tan(x) = (1-y)/(1+y)
= {tan (2n*pi+pi/4) - tan(z)}/{1+ tan(2n*pi+pi/4)*tan(z)}, where y = tanz
= tan (2n*pi+pi/4 - z)
Therefore:
x+z = (pi/4)*(8n+1)
= > x+ arctan(y) = (pi/4)*(8n+1)