Bryson Whitney

2022-04-07

I'm trying to prove the following problem:
$\frac{{\mathrm{sec}}^{2}\left(x\right)}{\mathrm{cot}\left(x\right)}-{\mathrm{tan}}^{3}\left(x\right)=\mathrm{tan}\left(x\right)$

Mey9ci0

We will work on the left side and show it is equl to $\mathrm{tan}\left(x\right)$. Let's change everything into $\mathrm{sin}\left(x\right)$ and $\mathrm{cos}\left(x\right)$ using the following formulas

${\mathrm{tan}}^{3}\left(x\right)=\frac{{\mathrm{sin}}^{3}\left(x\right)}{{\mathrm{cos}}^{3}\left(x\right)}$
This will give us:
$\frac{\frac{1}{{\mathrm{cos}}^{2}\left(x\right)}}{\frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}}-\frac{{\mathrm{sin}}^{3}\left(x\right)}{{\mathrm{cos}}^{3}\left(x\right)}$
Cleaning up the left fraction will give you
$\frac{\mathrm{sin}\left(x\right)}{{\mathrm{cos}}^{3}\left(x\right)}-\frac{{\mathrm{sin}}^{3}\left(x\right)}{{\mathrm{cos}}^{3}\left(x\right)}$
From here, factor out $\mathrm{sin}\left(x\right)$ from the top to get:
$\frac{\mathrm{sin}\left(x\right)\left(1-{\mathrm{sin}}^{2}\left(x\right)\right)}{{\mathrm{cos}}^{3}\left(x\right)}$
Finally, use the Pythagorean Identity ${\mathrm{cos}}^{2}\left(x\right)=1-{\mathrm{sin}}^{2}\left(x\right)$ to finish it off.

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