permaneceerc

2021-01-28

Verify the identity $\frac{\mathrm{cos}x}{1}-\mathrm{sin}x-\mathrm{tan}x=\frac{1}{\mathrm{cos}x}$

Elberte

$\frac{\mathrm{cos}\left(x\right)}{1-\mathrm{sin}\left(x\right)}-\mathrm{tan}\left(x\right)=$
$\frac{\mathrm{cos}\left(x\right)}{1-\mathrm{sin}\left(x\right)}-\frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}=$
$\left({\mathrm{cos}}^{2}\left(x\right)-\mathrm{sin}\left(x\right)\frac{1-\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)\left(1-\mathrm{sin}\left(x\right)\right)}=$
$\frac{{\mathrm{cos}}^{2}\left(x\right)-\mathrm{sin}\left(x\right)+{\mathrm{sin}}^{2}\left(x\right)}{\mathrm{cos}\left(x\right)\left(1-\mathrm{sin}\left(x\right)\right)}=$
$\frac{1-\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)\left(1-\mathrm{sin}\left(x\right)\right)}=$
$\frac{1}{\mathrm{cos}\left(x\right)}QED$

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