Differentiation over the hole PDE gives contradiction?

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I have two PDE: $$\bigg(\frac{1}{R_1}\partial_rR_2+K(r)\bigg)+\bigg(-\frac{1}{S_1}\partial_\theta S_2-T(\theta)\bigg)+A(r)\sin\theta\frac{S_2}{S_1}=0\,, \\ \bigg(-\frac{1}{R_2}\partial_rR_1-K(r)\bigg)+\bigg(\frac{1}{S_2}\partial_\theta S_1+T(\theta)\bigg)+A(r)\sin\theta\frac{S_1}{S_2}=0\,,$$ with the $R_1,R_2$ only functions of $(r)$ and the $S_1,S_2$ only functions of $\theta$.

If I differentiate the first equation by $\partial_r\partial_\theta$ it gives $\partial_rA\,\partial_\theta\bigg(\sin\theta\frac{S_2}{S_1}\bigg)=0$ so that one of the two terms is a constant. The same follows for the second equation.

$A(r)$ cannot be a constant since it would lead to a constraint in $r$ due to its form.

If the $\theta$ dependent term,(the second term) is constant then they lead to a contradiction: $$\lambda_1=\sin\theta\frac{S_2}{S_1}\,,\,\lambda_2=\sin\theta\frac{S_1}{S_2}\rightarrow \lambda_1\lambda_2=\sin^2\theta\,$$ that goes in contrary with the assumption that $\lambda_i$ is constant.

Does one cannot make such considerations over differentiating over a PDE?