[tex]\bf sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta)\\\\
sin(\theta)=\pm\sqrt{1-cos^2(\theta)}\\\\
cos(\theta)=\cfrac{1}{sec(\theta)}\\\\
-----------------------------\\\\
sin(t)\implies \pm \sqrt{1-cos^2(t)}\implies \pm \sqrt{1-\cfrac{1^2}{sec^2(t)}}\\\\\\ \pm \sqrt{\cfrac{sec^2(t)-1}{sec^2(t)}}[/tex]
so.. hmm which is it? the positive or the negative one?
well, we know angle "t" is in the IV quadrant, sine is negative on the IV quadrant, thus, is the negative one [tex]\bf -\sqrt{\cfrac{sec^2(t)-1}{sec^2(t)}}[/tex]
