Question

If $\tan A,\tan B$ are the roots of the quadratic $ab{x}^2-c^{2}x+ab=0$, where $a,b,c$ are the sides of a triangle, then prove
$\tan A=\frac{b}{a}$

Answer 1

$ab{x}^2-c^{2}x+ab=0$

$\tan A+\tan B=\frac{c^2}{ab}$

$\tan A\tan B=1$

$\tan (A+B)=\frac{\frac{c^2}{ab}}{1-1}=\infty$

$A+B=\frac{\pi }{2},c=\frac{\pi }{2}$

$\tan A=\frac{c^2+\sqrt{c^4-4a^2{b}^2}}{2ab}$

$\tan A=\frac{c^2-(a^2-b^2)}{2ab},c^2=a^2+b^2\therefore c^4=a^4+b^4+2a^2{b}^2$

$=\frac{a^2+b^2-a^2+b^2}{2ab}$

$=\frac{2b^2}{2ab}$

$\tan A=\frac{b}{a}$