The trigonometric functions sec(135^@)

To find the value of the trigonometric functions, first, convert sec function in terms of ${\displaystyle \cos \left(\theta \right)}$.
Use the identity ${\displaystyle \sec \left(\theta \right)=\frac{1}{\cos \left(\theta \right)}}$
We have
${\displaystyle \theta ={135}^{\circ }={90}^{\circ }+{45}^{\circ }}$
${\displaystyle \cos \left({135}^{\circ }\right)=\cos ({90}^{\circ }+{45}^{\circ })}$
${\displaystyle =-\sin \left({45}^{\circ }\right)\{\cos ({90}^{\circ }+\theta )=-\sin \left(\theta \right)}$
${\displaystyle =-\frac{1}{\sqrt{2}}}$
Now ${\displaystyle \sec \left(\theta \right)=\frac{1}{\cos \left(\theta \right)}}$
${\displaystyle \sec \left({135}^{\circ }\right)=\frac{1}{\cos \left({135}^{\circ }\right)}}$
${\displaystyle =\frac{1}{-\sin \left({45}^{\circ }\right)}}$
${\displaystyle =\frac{1}{-\frac{1}{\sqrt{2}}}}$
${\displaystyle =-\sqrt{2}}$
$=-1.4142$