Keenan Santos

2022-07-10

Proving $\frac{\mathrm{sin}x}{1-\mathrm{sin}x}-\frac{\mathrm{sin}x}{1+\mathrm{sin}x}\equiv 2{\mathrm{tan}}^{2}x$
What I have done so far is expanded them:
$\frac{\mathrm{sin}x\phantom{\rule{thickmathspace}{0ex}}\left(1+\mathrm{sin}x\right)}{\left(1-\mathrm{sin}x\right)\left(1+\mathrm{sin}x\right)}-\frac{\mathrm{sin}x\phantom{\rule{thickmathspace}{0ex}}\left(1-\mathrm{sin}x\right)}{\left(1+\mathrm{sin}x\right)\left(1-\mathrm{sin}x\right)}$
So therefore:
$\frac{\mathrm{sin}x+{\mathrm{sin}}^{2}x}{1-{\mathrm{sin}}^{2}x}-\frac{\mathrm{sin}x-{\mathrm{sin}}^{2}x}{1-{\mathrm{sin}}^{2}x}$
I'm completely stuck on what to do next

Karissa Macdonald

Hint: Simplify the original expression, You get ${\mathrm{cos}}^{2}\left(x\right)$ in denominator and $2{\mathrm{sin}}^{2}\left(x\right)$ in numerator:
By cross multiplying, you get
$\frac{\mathrm{sin}x+{\mathrm{sin}}^{2}x-\mathrm{sin}x+{\mathrm{sin}}^{2}x}{1-{\mathrm{sin}}^{2}x}=\frac{2{\mathrm{sin}}^{2}x}{1-{\mathrm{sin}}^{2}x}$
Now use
$1-{\mathrm{sin}}^{2}x={\mathrm{cos}}^{2}x$

Lorena Beard

You have
$\frac{\mathrm{sin}x}{1-\mathrm{sin}x}-\frac{\mathrm{sin}x}{1+\mathrm{sin}x}=\frac{\mathrm{sin}x+{\mathrm{sin}}^{2}x}{1-{\mathrm{sin}}^{2}x}-\frac{\mathrm{sin}x-{\mathrm{sin}}^{2}x}{1-{\mathrm{sin}}^{2}x}$
Then, use that $\frac{B}{A}-\frac{C}{A}=\frac{B-C}{A}$ to get
$\frac{\mathrm{sin}x+{\mathrm{sin}}^{2}x-\left(\mathrm{sin}x-{\mathrm{sin}}^{2}x\right)}{1-{\mathrm{sin}}^{2}x}=\frac{2{\mathrm{sin}}^{2}x}{1-{\mathrm{sin}}^{2}x}$
Now use $1-{\mathrm{sin}}^{2}x={\mathrm{cos}}^{2}x$

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