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?

Write $x=\tan A, y=\tan B, z=\tan C$.
By Vieta’s formula, we have
$$x+y+z=a,\ \ xy+yz+zx=0,\ \ xyz=-b.$$
Now you want to find the value of $$(1+x^2)(1+y^2)(1+z^2)=1+x^2+y^2+z^2+x^2y^2+y^2z^2+z^2x^2+x^2y^2z^2.$$
Note that
$$a^2=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx),$$
so $x^2+y^2+z^2=a^2$. Similarly,
$$0=(xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)$$
implies $x^2y^2+y^2z^2+z^2x^2=2ab$. Hence
$$(1+x^2)(1+y^2)(1+z^2)=1+a^2+2ab+b^2=1+(a+b)^2.$$
