# The trigonometric functions sec(135^@)

The trigonometric functions $\mathrm{sec}\left({135}^{\circ }\right)$
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d2saint0

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

Jeffrey Jordon