To find value of $\tan^2 C$

Question

In a $\triangle{ABC}$ , if $\tan A + 2\tan B=0$ to find range of values of $\tan^2 C$

I don't know how to begin

Thanks for help

Answer 1

We have $$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$

$$\implies-2\tan B+\tan B+\tan C=(-2\tan B)\tan B\tan C$$

$$\tan C\cdot2\tan^2B-\tan B+\tan C=0$$ which is a Quadratic Equation in $\tan B$

As $\tan B$ is real, the discriminant must be $\ge0$

Again, $\tan C=\dfrac{\tan B}{1+2\tan^2B}$

What if $B\to0$ or $\to\dfrac\pi2$

Answer 2

For every triangle $ABC$ we have $$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$

Then $$\tan C-\tan B=-2\tan^2 B\tan C$$ and, completing the square, $$2\tan C\left(\tan B-\frac1{4\tan C}\right)^2-\frac1{8\tan C}+\tan C=0$$

Hence $$\frac1{8\tan^2C}-1=2\left(\tan B-\frac1{4\tan C}\right)^2\ge 0$$

Answer 3

See if you can get these hints

$$ t_C= t_{\pi-A-B} = \dfrac{-t_A-t_B}{1-t_A t_B} =\dfrac{2t_B}{1+2 t_B^2}$$