Because $-\frac{\pi}{2} < -1 < 0$, the corresponding angle is in the $IV$ quadrant. Hence, $\sin(-1) < 0$, $\cos (-1) > 0$, and $\tan(-1) < 0$. Meanwhile, \t$$|\cos(-1) < 1\implies |\tan(-1)| = \Big|\frac{\sin(-1)}{\cos(-1)}\Big| >|\sin(-1)| $$ \t$$\therefore\quad \tan(-1) < \sin (-1) < \cos (-1)$$