Answer
$\sec$ -495$^{\circ}$ = -$\sqrt2$
Work Step by Step
$\csc$ -495$^{\circ}$
First, lets find the coterminal angle.
-495$^{\circ}$ + 2(360$^{\circ}$) = 225$^{\circ}$
Next, we find the reference angle.
$\theta$$^{1}$ = 225$^{\circ}$ - 180$^{\circ}$ = 45$^{\circ}$
Since the coterminal angle is in Quadrant III, $\csc$ is negative.
Therefore:
$\sec$ 495$^{\circ}$ = -$\sqrt2$
