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Consider the polar equation $r=2\csc\theta$. Rewriting this, we may say that $r\sin\theta=2$ or simply $y=2$. Is there not something lost along the way? It seems almost inaccurate to say $r=2\csc\theta\Leftrightarrow y=2$ from one system to the next (at least without some sort of remark) since $$ r=2\csc\theta\Leftrightarrow r=\frac{2}{\sin\theta}\Leftrightarrow r\sin\theta=2\Leftrightarrow y=2 $$ glosses over the fact that we cannot have $\theta=\pi k, k\in\mathbb{Z}$.

Is the reason we say this since $\lim_{\theta\to0^\pm}\frac{1}{\sin\theta}\to\pm\infty$ and hence $r$ extends infinitely far? Is there more rigorous justification needed for equality of equations in the two different systems?

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    You also need to exclude the interval $\theta\in (-\pi, 0)$, in which the cosecant is negative. And the repetitions of such interval due to periodicity. – 2017-01-15

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This is surly true that $r=2\csc\theta$ is equivalent to $y=2$, but more precise is we say $$y=2\,\,\,\equiv\,\,\,r=2\csc\theta\,\,,\theta\in\big(k\pi,(k+1)\pi\big)$$ This means when $\theta$ travel from $0$ to $\pi$, the points span the line $y=2$ from $+\infty$ to $-\infty$. This scenario is repeated for every $k$ to make a interval $\big(k\pi,(k+1)\pi\big)$ an span the line every time. This is because $\displaystyle\lim_{\theta\to0^+}\frac{2}{\sin \theta}\to+\infty$ and $\displaystyle\lim_{\theta\to\pi^-}\frac{2}{\sin \theta}\to-\infty$.