Why is $\frac{1}{1+(\frac{-\mathrm{sin}x}{1+\mathrm{cos}x}{)}^{2}}\equiv \frac{(1+\mathrm{cos}x{)}^{2}}{{\mathrm{sin}}^{2}x+1+2\mathrm{cos}x+{\mathrm{cos}}^{2}x}$?

Kaylynn Cook
2022-11-19
Answered

Why is $\frac{1}{1+(\frac{-\mathrm{sin}x}{1+\mathrm{cos}x}{)}^{2}}\equiv \frac{(1+\mathrm{cos}x{)}^{2}}{{\mathrm{sin}}^{2}x+1+2\mathrm{cos}x+{\mathrm{cos}}^{2}x}$?

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Kristen Garza

Answered 2022-11-20
Author has **13** answers

If you multiply the numerator and the denominator by $(1+\mathrm{cos}x{)}^{2}$, you get:

$\frac{1}{1+(\frac{-\mathrm{sin}x}{1+\mathrm{cos}x}{)}^{2}}=\frac{(1+\mathrm{cos}x{)}^{2}}{(1+\mathrm{cos}x{)}^{2}}\times \frac{1}{1+(\frac{-\mathrm{sin}x}{1+\mathrm{cos}x}{)}^{2}}=\frac{(1+\mathrm{cos}x{)}^{2}}{(1+\mathrm{cos}x{)}^{2}(1+(\frac{-\mathrm{sin}x}{1+\mathrm{cos}x}{)}^{2})}\phantom{\rule{0ex}{0ex}}=\frac{(1+\mathrm{cos}x{)}^{2}}{(1+\mathrm{cos}x{)}^{2}+{\mathrm{sin}}^{2}x}$

$\frac{1}{1+(\frac{-\mathrm{sin}x}{1+\mathrm{cos}x}{)}^{2}}=\frac{(1+\mathrm{cos}x{)}^{2}}{(1+\mathrm{cos}x{)}^{2}}\times \frac{1}{1+(\frac{-\mathrm{sin}x}{1+\mathrm{cos}x}{)}^{2}}=\frac{(1+\mathrm{cos}x{)}^{2}}{(1+\mathrm{cos}x{)}^{2}(1+(\frac{-\mathrm{sin}x}{1+\mathrm{cos}x}{)}^{2})}\phantom{\rule{0ex}{0ex}}=\frac{(1+\mathrm{cos}x{)}^{2}}{(1+\mathrm{cos}x{)}^{2}+{\mathrm{sin}}^{2}x}$

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