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

\(\displaystyle\theta={135}^{\circ}={90}^{\circ}+{45}^{\circ}\)

\(\displaystyle{\cos{{\left({135}^{\circ}\right)}}}={\cos{{\left({90}^{\circ}+{45}^{\circ}\right)}}}\)

\(\displaystyle=-{\sin{{\left({45}^{\circ}\right)}}}{\left\lbrace{\cos{{\left({90}^{\circ}+\theta\right)}}}=-{\sin{{\left(\theta\right)}}}\right.}\)

\(\displaystyle=-\frac{{1}}{\sqrt{{2}}}\)

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

Use the identity \(\displaystyle{\sec{{\left(\theta\right)}}}=\frac{{1}}{{{\cos{{\left(\theta\right)}}}}}\)

We have

\(\displaystyle\theta={135}^{\circ}={90}^{\circ}+{45}^{\circ}\)

\(\displaystyle{\cos{{\left({135}^{\circ}\right)}}}={\cos{{\left({90}^{\circ}+{45}^{\circ}\right)}}}\)

\(\displaystyle=-{\sin{{\left({45}^{\circ}\right)}}}{\left\lbrace{\cos{{\left({90}^{\circ}+\theta\right)}}}=-{\sin{{\left(\theta\right)}}}\right.}\)

\(\displaystyle=-\frac{{1}}{\sqrt{{2}}}\)

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