Can you prove that $\frac{a+b}{ab+1}$ is real if $|a|=1$, $|b|=1$, and $ab\ne-1$?

Question

Can you prove that $\dfrac{a+b}{ab+1}$ is real if $|a|=1$, $|b|=1$, and that $ab$ isn't equal to $-1$?

Answer 1

$$\dfrac{a+b}{ab+1}= \dfrac{(a+b)(1+\overline{a}\overline{b})}{(ab+1)(1+\overline{a}\overline{b})}=\dfrac{(a+b)(1+\overline{a}\overline{b})}{|ab+1|^2}=\dfrac{a+b+|a|^2\overline{b}+\overline{a}|b|^2}{|ab+1|^2}=\dfrac{a+b+\overline{b}+\overline{a}}{|ab+1|^2}=\dfrac{2Re(a+b)}{|ab+1|^2} \in \Bbb R$$

Answer 2

$ \frac{i \sin (x)+\cos (x)+i \sin (y)+\cos (y)}{1+(\cos (x)+i \sin (x)) (\cos (y)+i \sin (y))} = \cos \left(\frac{x-y}{2}\right) \sec \left(\frac{x+y}{2}\right) $