Gerald Ritter

2022-01-28

Prove that $\frac{1+2\mathrm{sin}\theta \mathrm{cos}\theta }{{\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta }=\frac{1+\mathrm{tan}\theta }{1-\mathrm{tan}\theta }$

Howard Gallagher

Expert

Note that
$1+2\mathrm{sin}\theta \mathrm{cos}\theta ={\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta +2\mathrm{sin}\theta \mathrm{cos}\theta ={\left(\mathrm{cos}\theta +\mathrm{sin}\theta \right)}^{2}$
Therefore
$\frac{1+2\mathrm{sin}\theta \mathrm{cos}\theta }{{\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta }=\frac{{\left(\mathrm{cos}\theta +\mathrm{sin}\theta \right)}^{2}}{\left(\mathrm{cos}\theta +\mathrm{sin}\theta \right)\left(\mathrm{cos}\theta -\mathrm{sin}\theta \right)}$
$=\frac{\mathrm{cos}\theta +\mathrm{sin}\theta }{\mathrm{cos}\theta -\mathrm{sin}\theta }$
$=\frac{\mathrm{cos}\theta \left(1+\mathrm{tan}\theta \right)}{\mathrm{cos}\theta \left(1-\mathrm{tan}\theta \right)}$
$=\frac{1+\mathrm{tan}\theta }{1-\mathrm{tan}\theta }$

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