When constructing an orthogonal matrix with a rational parametrisation, we can simply deduce:
$$Q(t) = \frac{1}{1+t^2} \begin{pmatrix}1-t^2 & -2t\\ 2t & 1-t^2\end{pmatrix}.$$
There is a way to write a product of two orthogonal matrices in this form, which we'll say is $Q(r)Q(t)$. With this, we'd have to find some value $u$ such that $Q(r)Q(t) = Q(u)$. What would be this value of $u$ and how would you be able to find it?
Make the substitution $t = \tan \frac \theta 2$.
We see that $Q(\tan\frac \theta 2)=\begin{pmatrix} \cos \theta & -\sin\theta \\ \sin \theta & \cos \theta\end{pmatrix}$ is simply a rotation of angle $\theta$.
So let us write $Q(r)Q(t)$ as $Q(\tan\frac\alpha 2)Q(\tan\frac \beta2)$ for some $\alpha, \beta \in (-\pi, \pi) $.
This is a composition of two rotations of angles $\alpha$ and $\beta$ - which is just a rotation of angle $\alpha + \beta$.
Thus, $$\begin{aligned}Q(u)= Q(r)Q(t)&=Q\left(\tan\frac\alpha 2\right)Q\left(\tan\frac \beta2\right)\\&= Q\left(\tan \frac{\alpha + \beta}2\right), \end{aligned}$$
so $u = \tan \frac{\alpha+\beta}{2}.$
Now just apply the tangent angle sum identity:
$$\tan \frac{\alpha+\beta}{2} = \frac{\tan\frac\alpha 2+\tan\frac\beta 2}{1-\tan\frac\alpha 2\tan\frac\beta 2}.$$
We thus have $$u = \frac{r+t}{1-rt}.$$