<s>[INST] If $\tan A, \tan B, \tan C$ are roots of $x^3-ax^2+b=0$, find $(1+\tan^2 A)(1+\tan^2B)(1+\tan^2C)$ Here are all the results I got
$$\tan(A+B+C)=a-b$$
And
$$(1+\tan^2A)(1+\tan^2B)(1+\tan^2C)=(\frac{1}{\cos A\cos B\cos C})^2$$
And
$$\cot A+\cot B + \cot C=0$$
How should I use these results?
 [/INST] Set $x = 1 + y^2 \Rightarrow y = ±\sqrt{x - 1}$ so that $1 + \tan^2 A, 1 + \tan^2 B, 1 + \tan^2 C$ give the roots of $x^3 - ax^2 + b = 0$.
This gives $±(x-1)^{3/2} - a(x-1) +b=0$. By Vieta, the product of the roots is the constant term, but we do not have a polynomial yet. Squaring both sides of $±(x  - 1)^{3/2} = a(x-1) - b$, the constant term is $(-1)^3 - (-a-b)^2 = -1 -(a+b)^2$ if we take the positive branch, and $(-a-b)^2 - (-1) = 1 + (a+b)^2$ if we take the negative branch. However, since $(1 + \tan^2 A)(1 + \tan^2 B)(1 + \tan^2 C)$ is always positive, the answer is $1 + (a+b)^2$.
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