# Prove that: 1+cosx/1-cosx=tan^2x/(secx-1)^2

Prove that: $1+\frac{\mathrm{cos}x}{1}-\mathrm{cos}x=\frac{{\mathrm{tan}}^{2}x}{{\left(\mathrm{sec}x-1\right)}^{2}}$
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Alix Ortiz
$\frac{1+\mathrm{cos}\left(x\right)}{1-\mathrm{cos}\left(x\right)}=\frac{1-{\mathrm{cos}}^{2}\left(x\right)}{{\left(1-\mathrm{cos}\left(x\right)\right)}^{2}}=$
$\frac{{\mathrm{sin}}^{2}\left(x\right)}{{\mathrm{cos}}^{2}\left(x\right)}{\left(\mathrm{sec}\left(x\right)-1\right)}^{2}=$
$\frac{{\mathrm{tan}}^{2}\left(x\right)}{{\left(\mathrm{sec}\left(x\right)-1\right)}^{2}}$