The Weierstraß $\wp$ function for a lattice $\Lambda\subset\Bbb C$ can be defined by the sum
$$ \wp(z) = \frac1{z^2} ~+\!\! \sum_{\lambda\in\Lambda\setminus\{0\}} \left(\frac1{(z-\lambda)^2}-\frac1{\lambda^2}\right) \tag1 $$ and there is the following representation of $\csc$: $$ f(x) := \pi^2\csc^2(\pi x) = \sum_{n\in\Bbb Z}\frac1{(x+n)^2} \tag 2 $$
My question is if (and how) $\wp$ can be expressed as a "one-dimensional" sum over $f(x)$ or other trigonometric functions. For simplicity, let's assume that $\Lambda=\Bbb Z + \Bbb Zi$.
Splitting the sum $(1)$ into two sums, $A$ over $\lambda\in\Bbb R$, and $B$ over $\lambda\in\Bbb C\setminus \Bbb R$:
$$ \wp(z) = \underbrace{\frac1{z^2} ~+\!\! \sum_{n\in\Bbb Z\setminus\{0\}} \left(\frac1{(z-n)^2}-\frac1{n^2}\right)}_{\textstyle=:A(z)} ~+\!\!\underbrace{\sum_{m\in\Bbb Z\setminus\{0\}} \sum_{n\in\Bbb Z} \left(\frac1{(z-n-mi)^2}-\frac1{(n+mi)^2}\right)}_{\textstyle=:B(z)} \tag 3 $$ This should be correct because $(1)$ converges absolutely and thus the series can be rearranged. The two summands of $A$ also converge absolutely, and therefore we have:
$$ A(z) = f(z) - 2\zeta(2) \tag 4 $$ Sum $B$ is very similar, but there's an additional sum in the imaginary direction over $f$:
$$\begin{align} B(z) &=\sum_{m\in\Bbb Z\setminus\{0\}} \left(f(z-mi) -\sum_{n\in\Bbb Z} \frac1{(n+mi)^2}\right) \tag{5.1}\\ &=\sum_{m\in\Bbb Z\setminus\{0\}} \bigl(f(z-mi) - f(mi)\bigr) \tag{5.2}\\ &=\sum_{m\in\Bbb Z\setminus\{0\}} f(z-mi) - \underbrace{\sum_{m\in\Bbb Z\setminus\{0\}} f(mi)}_{\textstyle =:C} \tag{5.3} \end{align}$$
Taking it all together, it's $$ \wp(z) =A(z)+B(z) = -2\zeta(2) -C +\sum_{m\in\Bbb Z} f(z-mi) \tag 6 $$
But then I am stuck to determine $C$.
