Question

If $\tan A,\tan B$ are the roots of quadratic equation $a{x}^2+3x+4=0,a\ne 0$ and $\frac{\tan A}{\cot B-\tan A}=\frac{1}{3}$, then the value of $a$ is

Answer 1

$\because \tan A$and $\tan B$ are the roots of $a{x}^2+3x+4=0.$
$\Rightarrow \tan A\tan B=\frac{4}{a}$
So, $\frac{\tan A}{\cot B-\tan A}=\frac{1}{3}$
$\Rightarrow \frac{\tan A\tan B}{1-\tan A\tan B}=\frac{1}{3}$
$\Rightarrow \frac{\frac{4}{a}}{1-\frac{4}{a}}=\frac{1}{3}$
$\Rightarrow \frac{4}{a-4}=\frac{1}{3}\Rightarrow a=16$