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By doing
$\sqrt{-\frac{1}{f\left(r\right)^{2} \sin\left(r\right)^{2}}} - \frac{1}{\sqrt{-f\left(r\right)^{2} \sin\left(r\right)^{2}}} = \frac{i}{\sqrt{f\left(r\right)^{2} \sin\left(r\right)^{2}}}-\frac{1}{i\sqrt{f\left(r\right)^{2} \sin\left(r\right)^{2}}} = \frac{2i}{\sqrt{f\left(r\right)^{2} \sin\left(r\right)^{2}}}$
I wouldn't expect bool(eq==0) to return True.
You can deal with square roots using canonicalize_radical().
eq.canonicalize_radical()
returns $\frac{2 i}{f\left(r\right) \sin\left(r\right)}$.
