Answer
$-\sqrt 2$
Work Step by Step
The reference angle of an angle $0 \leq \theta \lt 2\pi $ based on its position can be computed by using the following steps:
a) Quadrant- I: $\theta $
b) Quadrant- II: $180^{\circ}-\theta $
c) Quadrant -III: $\theta - 180^o $
d) Quadrant -IV: $360^{\circ}-\theta $
Reference angle of $495^{\circ}$ is equal to
$ =540^{\circ}-495^{\circ}=45^{\circ}$
Since, $\sec 45^{\circ}=\sqrt 2$
So, $\sec (495)^{\circ}=-\sqrt 2$; Because $\theta $ lies in Quadrant-II.
