pierdoodsu

2021-12-31

Here are all the results I got
$\mathrm{tan}\left(A+B+C\right)=a-b$
And
$\left(1+{\mathrm{tan}}^{2}A\right)\left(1+{\mathrm{tan}}^{2}B\right)\left(1+{\mathrm{tan}}^{2}C\right)={\left(\frac{1}{\mathrm{cos}A\mathrm{cos}B\mathrm{cos}C}\right)}^{2}$
And
$\mathrm{cot}A+\mathrm{cot}B+\mathrm{cot}C=0$
How should I use these results?

veiga34

Expert

Write $x=\mathrm{tan}A,y=\mathrm{tan}B,z=\mathrm{tan}C$
By Vieta’s formula, we have

Now you want to find the value of
$\left(1+{x}^{2}\right)\left(1+{y}^{2}\right)\left(1+{z}^{2}\right)=1+{x}^{2}+{y}^{2}+{z}^{2}+{x}^{2}{y}^{2}+{y}^{2}{z}^{2}+{z}^{2}{x}^{2}+{x}^{2}{y}^{2}{z}^{2}$
Note that
${a}^{2}={\left(x+y+z\right)}^{2}={x}^{2}+{y}^{2}+{z}^{2}+2\left(xy+yz+zx\right)$
so, ${x}^{2}+{y}^{2}+{z}^{2}={a}^{2}$. Similarly,
$0={\left(xy+yz+zx\right)}^{2}={x}^{2}{y}^{2}+{y}^{2}{z}^{2}+{z}^{2}{x}^{2}+2xyz\left(x+y+z\right)$
implies ${x}^{2}{y}^{2}+{y}^{2}{z}^{2}+{z}^{2}{x}^{2}=2ab$. Hence
$\left(1+{x}^{2}\right)\left(1+{y}^{2}\right)\left(1+{z}^{2}\right)=1+{a}^{2}+2ab+{b}^{2}=1+{\left(a+b\right)}^{2}$

lovagwb

Expert

Let $p\left(x\right)={x}^{3}-a{x}^{2}+b$, then
$p\left(x\right)=\left(x-\mathrm{tan}A\right)\left(x-\mathrm{tan}B\right)\left(x-\mathrm{tan}C\right)$
Now you need
$p\left(i\right)\cdot p\left(-i\right)$

Vasquez

Expert

Let $y={x}^{2}+1$ so that ${x}^{2}=y-1$ and substitute as much as possible in the original cubic:
$x\left(y-1\right)-a\left(y-1\right)+b=0$
Rearrange to isolate x and obtain:
$x\left(y-1\right)=a\left(y-1\right)+b=ay-\left(a+b\right)$
Now square this:
${x}^{2}\left(y-1{\right)}^{2}=\left(y-1{\right)}^{3}=\left(ay-\left(a+b\right){\right)}^{2}$
so that
${y}^{3}-3{y}^{2}+3y+1-{a}^{2}{y}^{2}+2a\left(a+b\right)y-\left(a+b{\right)}^{2}=0$

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