Answer
$-\sqrt {2}$
Work Step by Step
$\sec(-495^{\circ})=\frac{1}{\cos (-495^{\circ})}$
Let us find $\cos (-495^{\circ})$ first.
As the period of cosine function is $360^{\circ}$,
$\cos(-495^{\circ})=\cos(-495^{\circ}+360^{\circ})$
$=\cos(-135^{\circ})=\cos 135^{\circ}$
(As $\cos (-x)=\cos x$ )
$=\cos(180^{\circ}-45^{\circ})=-\cos 45^{\circ}$
( As $\cos(180^{\circ}-x)=-\cos x$ )
$=-\frac{1}{\sqrt 2}$
$\sec(-495^{\circ})=\frac{1}{-\frac{1}{\sqrt 2}}=-\sqrt 2$
