Note that I take the Definition of sl (Lemniscate sine) and cl (Lemniscate cosine) as the inverse of $\int_0^z \frac{1}{\sqrt{1-t^4}} dt$ and $\int_z^1 \frac{1}{\sqrt{1-t^4}} dt$.
Following from this post, I am able to go from the definition to the identities $$\text{cl}^2(t)+\text{sl}^2(t)+\text{cl}^2(t)\text{sl}^2(t)=1$$ $$\text{cl}^2(t)=\frac{1-\text{sl}^2(t)}{1+\text{sl}^2(t)} , \text{sl}^2(t)=\frac{1-\text{cl}^2(t)}{1+\text{cl}^2(t)}$$
$$\text{sl}(u+v)=\frac{\text{sl}(u) \text{sl}'(v)+ \text{sl}(v) \text{sl}'(u)}{1+\text{sl}^2(u) \text{sl}^2(v)}$$ $$\text{cl}(x+y)=\frac{\text{cl}(x) \text{sl}'(y)+ \text{sl}(y) \text{cl}'(x)}{1+\text{cl}^2(x) \text{sl}^2(y)}$$ However, I wasn't able to go from here to the formula on Wikipedia
$$\text{sl}(u+v)=\frac{\text{sl}(u) \text{cl}(v)+ \text{sl}(v) \text{cl}(u)}{1-\text{sl}(u) \text{sl}(v)\text{cl}(u) \text{cl}(v)}$$
I tried using that sl$'(t)=(1+\text{sl}^2(t))\text{cl}(t)$ and cl$'(t)=-(1+\text{cl}^2(t))\text{sl}(t)$ but it did not work
Any help is appreciated.
