Haialarmz6

2022-01-26

Proving ${\mathrm{tan}}^{2}\left(\frac{\pi }{4}+x\right)=\frac{1+\mathrm{sin}\left(2x\right)}{1-\mathrm{sin}\left(2x\right)}$

fionaluvsyou0x

Expert

${\mathrm{tan}}^{2}\left(\frac{\pi }{4}+x\right)$
$={\left(\frac{\mathrm{tan}\frac{\pi }{4}+\mathrm{tan}x}{1-\mathrm{tan}\frac{\pi }{4}\cdot \mathrm{tan}x}\right)}^{2}$
$=\frac{{\left(1+\mathrm{tan}x\right)}^{2}}{{\left(1-\mathrm{tan}x\right)}^{2}}$
$=\frac{{\left(1+\frac{\mathrm{sin}x}{\mathrm{cos}x}\right)}^{2}}{{\left(1-\frac{\mathrm{sin}x}{\mathrm{cos}x}\right)}^{2}}$
$=\frac{{\left(\mathrm{cos}x+\mathrm{sin}x\right)}^{2}}{{\left(\mathrm{cos}x-\mathrm{sin}x\right)}^{2}}$
$=\frac{{\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x+2\mathrm{sin}x\mathrm{cos}x}{{\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x-2\mathrm{sin}x\mathrm{cos}x}$
$=\frac{1+2\mathrm{sin}x\mathrm{cos}x}{1-2\mathrm{sin}x\mathrm{cos}x}$
$=\frac{1+\mathrm{sin}2x}{1-\mathrm{sin}2x}$

sphwngzt

Expert

$1±\mathrm{sin}2x={\left(\mathrm{cos}x±\mathrm{sin}x\right)}^{2}$
$\frac{\mathrm{cos}x+\mathrm{sin}x}{\mathrm{cos}x+\mathrm{sin}x}=\frac{1+\mathrm{tan}x}{1-\mathrm{tan}x}=\mathrm{tan}\left(x+\frac{\pi }{4}\right)$

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