Question

A, B, C are the angles of a triangle such that $\tan A,\tan B,\tan C$ are three roots of the biquadratic $x^4-p{x}^3+q{x}^2-rx+s=0$. Prove that the fourth root satisfies $x^2-px+s=0$.

Answer 1

Let the fourth root be $\alpha$
$\therefore \tan A+\tan B+\tan C+\alpha =p$ ..$\left(1\right)$
$\left(\tan A\tan B\tan C\right)\alpha =s$ ..$\left(2\right)$
Also we know that in a triangle ABC $\tan A+\tan B+\tan C=\tan A\tan B\tan C$ $\left(3\right)$
$\therefore p-\alpha =\frac{s}{\alpha }$ or ${\alpha }^2-p\alpha +s=0$
$\therefore \alpha$ is a root of $x^2-px+s=0$.