Rivka Thorpe

2021-09-12

Proof trigonometric identities.
$\mathrm{sec}\left(x\right)+\mathrm{tan}\left(x\right)=\frac{\mathrm{cos}\left(x\right)}{1-\mathrm{sin}\left(x\right)}$

Ayesha Gomez

The given equation is $\mathrm{sec}\left(x\right)+\mathrm{tan}\left(x\right)=\frac{\mathrm{cos}\left(x\right)}{1-\mathrm{sin}\left(x\right)}$
Let choose $\mathrm{sec}\left(x\right)+\mathrm{tan}\left(x\right)$ to prove $\frac{\mathrm{cos}x}{1-\mathrm{sin}x}$ to prove the identity.
$\mathrm{sec}x+\mathrm{tan}x=\frac{1}{\mathrm{cos}x}+\frac{\mathrm{sin}x}{\mathrm{cos}x}$
($\because \mathrm{sec}x=\frac{1}{\mathrm{cos}x},\mathrm{tan}x=\frac{\mathrm{sin}x}{\mathrm{cos}x}$)
$=\frac{1+\mathrm{sin}x}{\mathrm{cos}x}$
$=\frac{1+\mathrm{sin}x}{\mathrm{cos}x}\cdot \frac{\mathrm{cos}x}{\mathrm{cos}x}$
$=\frac{\left(1+\mathrm{sin}x\right)\mathrm{cos}x}{{\mathrm{cos}}^{2}x}$
On further simplification,
$\mathrm{sec}x+\mathrm{tan}x=\frac{\left(1+\mathrm{sin}x\right)\mathrm{cos}x}{\left(1-{\mathrm{sin}}^{2}x\right)}$
$=\frac{\left(1+\mathrm{sin}x\right)\mathrm{cos}x}{{1}^{2}-{\mathrm{sin}}^{2}x}$
$=\frac{\left(1+\mathrm{sin}x\right)\mathrm{cos}x}{\left(1+\mathrm{sin}x\right)\left(1-\mathrm{sin}x\right)}$
$=\frac{\mathrm{cos}x}{\left(1-\mathrm{sin}x\right)}$
Hence, the given identity is proved.

Jeffrey Jordon