# Proof trigonometric identities.\frac{1-\sin(-x)}{\cos x+\cos(-x)\sin x}=\sec x

Proof trigonometric identities.
$\frac{1-\mathrm{sin}\left(-x\right)}{\mathrm{cos}x+\mathrm{cos}\left(-x\right)\mathrm{sin}x}=\mathrm{sec}x$
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Given :
$\frac{1-\mathrm{sin}\left(-x\right)}{\mathrm{cos}x+\mathrm{cos}\left(-x\right)\mathrm{sin}x}=\mathrm{sec}x$
Considering left-hand side $\frac{1-\mathrm{sin}\left(-x\right)}{\mathrm{cos}x+\mathrm{cos}\left(-x\right)\mathrm{sin}x}$
Using $\mathrm{sin}\left(-\theta \right)=-\mathrm{sin}\theta$ and $\mathrm{cos}\left(-\theta \right)=\mathrm{cos}\theta$,
$\frac{1-\mathrm{sin}\left(-x\right)}{\mathrm{cos}x+\mathrm{cos}\left(-x\right)\mathrm{sin}x}=\frac{1-\left(-\mathrm{sin}\left(x\right)\right)}{\mathrm{cos}x+\mathrm{cos}\left(x\right)\mathrm{sin}x}$
$\frac{1-\mathrm{sin}\left(-x\right)}{\mathrm{cos}x+\mathrm{cos}\left(-x\right)\mathrm{sin}x}=\frac{1+\mathrm{sin}\left(x\right)}{\mathrm{cos}x+\mathrm{cos}\left(x\right)\mathrm{sin}x}$
$\frac{1-\mathrm{sin}\left(-x\right)}{\mathrm{cos}x+\mathrm{cos}\left(-x\right)\mathrm{sin}x}=\frac{1+\mathrm{sin}\left(x\right)}{\mathrm{cos}x\left(1+\mathrm{sin}x\right)}$
$\frac{1-\mathrm{sin}\left(-x\right)}{\mathrm{cos}x+\mathrm{cos}\left(-x\right)\mathrm{sin}x}=\frac{1}{\mathrm{cos}x}$
Using $\mathrm{sec}\theta =\frac{1}{\mathrm{cos}\theta }$
$\frac{1-\mathrm{sin}\left(-x\right)}{\mathrm{cos}x+\mathrm{cos}\left(-x\right)\mathrm{sin}x}=\mathrm{sec}x$
Jeffrey Jordon