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?

It is basically an algebraic problem:
If $p,q,r$ are the roots of $$x^3-ax^2+b=0$$
it is sufficient to find the cubic equation whose roots are $w=p^2+1,y=q^2+1,z=r^2+1$
From $ap^2-b=p^3,$ we have
$$(ap^2-b)^2=(p^3)^2=(p^2)^3$$
Replace $p^2$ with $w-1$  to find $$(a(w-1)-b)^2=(w-1)^3$$
$$\iff w^3+(\cdots)w^2+(\cdots)w-1-(a+b)^2=0$$
Obviously,  we shall reach at the same equation if start with $y=q^2+1$ or $z=r^2+1$
Using Vieta's  formula, $$wyz=-\dfrac{-1-(a+b)^2}1$$
See also :

*

*Finding $\sum_{k=0}^{n-1}\frac{\alpha_k}{2-\alpha_k}$, where $\alpha_k$ are the $n$-th roots of unity


*If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, prove the following.
