Solving trigonometric equation $\arctan y^2+\arctan x^2= π/4$

Question

Given the equation $\arctan y^2+\arctan x^2= π/4$, I am asked to find $y$ given $x = 1/\sqrt2$. Even after reading the solution I'm still unsure of how to do it. In the solution, the following steps are displayed. I understand how the first line is obtained, but I don't understand how it goes from the first to the second. Where does the denominator come from? Should the first line not simply expand to the numerator of the second line?

Answer 1

HINT:

$$\tan(A\pm B)=\frac{\tan(A)\pm \tan(B)}{1\mp\tan(A)\tan(B)}$$

Answer 2

Using the formula $tan^{-1}A+tan^{-1}B= tan^{-1}\frac{A+B}{1-AB}$ we have, $tan^{-1}\frac{x^2+y^2}{1-x^2y^2}=\frac{\pi}{4}$ $\implies$$\frac{x^2+y^2}{1-x^2y^2}= tan\frac{\pi}{4}$ Can you solve it from here?