# Why is 1/(1+((-sin x)/(1+cos x))^2) -= ((1+cos x)^2)/(sin^2 x+1+2 cos x+cos^2 x)?

Why is $\frac{1}{1+\left(\frac{-\mathrm{sin}x}{1+\mathrm{cos}x}{\right)}^{2}}\equiv \frac{\left(1+\mathrm{cos}x{\right)}^{2}}{{\mathrm{sin}}^{2}x+1+2\mathrm{cos}x+{\mathrm{cos}}^{2}x}$?
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Kristen Garza
If you multiply the numerator and the denominator by $\left(1+\mathrm{cos}x{\right)}^{2}$, you get:
$\frac{1}{1+\left(\frac{-\mathrm{sin}x}{1+\mathrm{cos}x}{\right)}^{2}}=\frac{\left(1+\mathrm{cos}x{\right)}^{2}}{\left(1+\mathrm{cos}x{\right)}^{2}}×\frac{1}{1+\left(\frac{-\mathrm{sin}x}{1+\mathrm{cos}x}{\right)}^{2}}=\frac{\left(1+\mathrm{cos}x{\right)}^{2}}{\left(1+\mathrm{cos}x{\right)}^{2}\left(1+\left(\frac{-\mathrm{sin}x}{1+\mathrm{cos}x}{\right)}^{2}\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(1+\mathrm{cos}x{\right)}^{2}}{\left(1+\mathrm{cos}x{\right)}^{2}+{\mathrm{sin}}^{2}x}$