Averie Ray

2022-04-22

If $\mathrm{tan}x-\mathrm{sin}x=a$ , what is ${\mathrm{sec}}^{3}\left(x\right)-1$ in terms of a

Yaretzi Odom

Beginner2022-04-23Added 16 answers

We can recall the difference of cubes formula

${a}^{3}-{b}^{3}=(a-b)({a}^{2}+2ab+{b}^{2})$

So now we see that

${\mathrm{sec}}^{3}x-1=(\mathrm{sec}x-1)({\mathrm{sec}}^{2}x-2\mathrm{sec}x+1)$

Or that

${\mathrm{sec}}^{3}x-1=\left(\frac{A}{\mathrm{sin}x}\right)({\mathrm{sec}}^{2}x-2\mathrm{sec}x+1)$

Continue with the simplification process.

${\mathrm{sec}}^{3}x-1=\frac{A}{\mathrm{sin}x}(2-2\mathrm{sec}x+{\mathrm{tan}}^{2}x)$

${\mathrm{sec}}^{3}x-1=\frac{A}{\mathrm{sin}x}(-\frac{A}{\mathrm{sin}x}+{\mathrm{tan}}^{2}x)$

$\mathrm{sec}}^{3}x-1=\frac{A\mathrm{sin}x}{{\mathrm{cos}}^{2}x}-\frac{{A}^{2}}{{\mathrm{sin}}^{2}x$

We can't go much further than that, unfortunately. I would like to see if any contributors have other ideas.

So now we see that

Or that

Continue with the simplification process.

We can't go much further than that, unfortunately. I would like to see if any contributors have other ideas.

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