Correct answer: [-1,1]
Since sqrt(1+tan^2 x)=1/|cos x| and sqrt(1+cot^2 x)=1/|sin x|, f(x)=sin x|cos x| - cos x|sin x|. In quadrant II this equals -2 sin x cos x in (0,1], in quadrant IV it is in [-1,0), and 0 at the axes, so the value set is [-1, 1].