Bergen

2020-10-31

Solve the equation $\frac{1+{\mathrm{tan}}^{2}A}{1+{\mathrm{cot}}^{2}A}={\left(\frac{1-\mathrm{tan}A}{1-\mathrm{cot}A}\right)}^{2}={\mathrm{tan}}^{2}A$

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Work with the left side
$\frac{1+{\mathrm{tan}}^{2}A}{1+{\mathrm{cot}}^{2}A}$

$=\frac{\frac{{\mathrm{cos}}^{2}A+{\mathrm{sin}}^{2}A}{{\mathrm{cos}}^{2}A}}{\frac{{\mathrm{sin}}^{2}A+{\mathrm{cos}}^{2}A}{{\mathrm{sin}}^{2}A}}$
$=\frac{\frac{1}{{\mathrm{cos}}^{2}A}}{\frac{1}{{\mathrm{sin}}^{2}A}}\left[\because {\mathrm{sin}}^{2}A+{\mathrm{cos}}^{2}A=1\right]$
$=\frac{{\mathrm{sin}}^{2}A}{{\mathrm{cos}}^{2}A}={\mathrm{tan}}^{2}A$
Now note that
$\left(\left(1-\mathrm{tan}A\right)/\left(1-\mathrm{cot}A\right){\right)}^{2}$

$={\left(\frac{\frac{\mathrm{cos}A-\mathrm{sin}A}{\mathrm{cos}A}}{\frac{\mathrm{sin}A-\mathrm{cos}A}{\mathrm{sin}A}}\right)}^{2}$
$=\left(\left(-\left(sinA-cosA\right)\right)/cosA\ast sinA/\left(sinA-cosA\right){\right)}^{2}$
$={\left(-\frac{\mathrm{sin}A}{\mathrm{cos}A}\right)}^{2}=\frac{{\mathrm{sin}}^{2}A}{{\mathrm{cos}}^{2}A}={\mathrm{tan}}^{2}A$

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