In ${\Delta }^{le}ABC,$ $3TanA+4=0$  then the  quadratic equation whose roots are $\sec A$  and $co\sec A$ is

$TanA=\frac{-4}{3}$  then sec A = $-\frac{5}{3}$

            Cosec A = $\frac{5}{4}$

            $\therefore x^2-\left(\frac{-5}{3}+\frac{5}{4}\right)x+\left(\frac{-5}{3}\right).\frac{5}{4}=0$

                $\Rightarrow 12x^2+5x-25=0$