nicekikah

2021-02-19

Solve the equation $\frac{\mathrm{cot}\left(x\right)-\mathrm{tan}\left(x\right)}{\left({\mathrm{cot}}^{2}\left(x\right)-{\mathrm{tan}}^{2}\left(x\right)}=\mathrm{sin}x\mathrm{cos}x$

Ezra Herbert

Apply Difference of Two Squares Formula:
${x}^{2}-{y}^{2}=\left(x+y\right)\left(x-y\right)$
$\to {\mathrm{cot}}^{2}\left(x\right)-{\mathrm{tan}}^{2}\left(x\right)=\left(\mathrm{cot}\left(x\right)+\mathrm{tan}\left(x\right)\right)\left(\mathrm{cot}\left(x\right)-\mathrm{tan}\left(x\right)\right)$
$\phantom{\rule{0.278em}{0ex}}⟹\phantom{\rule{0.278em}{0ex}}\frac{\mathrm{cot}\left(x\right)-\mathrm{tan}\left(x\right)}{{\mathrm{cot}}^{2}\left(x\right)-{\mathrm{tan}}^{2}\left(x\right)}=\frac{\mathrm{cot}\left(x\right)-\mathrm{tan}\left(x\right)}{\left(\mathrm{cot}\left(x\right)+\mathrm{tan}\left(x\right)\right)\left(\mathrm{cot}\left(x\right)-\mathrm{tan}\left(x\right)\right)}=\frac{1}{\left(\mathrm{cot}\left(x\right)+\mathrm{tan}\left(x\right)}$
$=\frac{1}{2\mathrm{csc}\left(2x\right)}$
$=\frac{\mathrm{sin}\left(2x\right)}{2}$
$=\mathrm{sin}x\mathrm{cos}x$
Therefore
$\sqrt{{a}^{2}-{a}^{2}{\mathrm{sin}}^{2}\left(\theta \right)}=a\mathrm{cos}\left(\theta \right)$

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