2
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Mathematica correctly simplifies the following expression:

Assuming[-Pi <= θ <= Pi, FullSimplify[TrigToExp[
                                    1/Sqrt[1 + Abs[Cot[θ/2]]^2] - Abs[Sin[θ/2]]]]]
(*0*)

But it will not simplify

Assuming[-Pi <= θ <= Pi, FullSimplify[1/Sqrt[1 + Abs[Cot[θ/2]]^2]]]
(*1/Sqrt[1 + Abs[Cot[θ/2]]^2]*)

to

Abs[Sin[θ/2]]

Even though it has less fractions, less square-roots and less sums.

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3
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It works if you exclude the singular point:

Assuming[2 Pi > θ > 0,  FullSimplify[1/Sqrt[1 + Abs[Cot[θ/2]]^2]]]
(* Sin[θ/2] *)
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  • $\begingroup$ But why is Mathematica able to deal with the singular point in my first FullSimplify expression? $\endgroup$ – NOhs Aug 18 '15 at 14:16
  • $\begingroup$ @MrZ Because TrueQ[ComplexInfinity == ComplexInfinity] is false. Try Reduce[1/Sqrt[1 + Abs[Cot[\[Theta]/2]]^2] == Sin[\[Theta]/2], \[Theta], Reals] /. C[1] -> 0 $\endgroup$ – Dr. belisarius Aug 18 '15 at 14:24

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