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

• A. $\tan A=\frac{a}{b}$
• B. $\tan B=\frac{b}{a}$
• C. $\cos C=0$
• D. $\tan A+\tan A=\frac{c^2}{ab}$

Answer 1

Answer: (A), (B), (D)

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

If $\tan A,\tan B$ are the roots of the quadratic $ab{x}^2-c^{2}x+ab=0$ then
$\tan A+\tan B=\frac{c^2}{ab}$
$\tan A\tan B=1$
$\because \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{\frac{c^2}{ab}}{1-1}=\frac{c^2}{0}=\infty$
$\therefore A+B=\frac{\pi }{2}$
$\therefore C=\frac{\pi }{2}$
$\therefore \cos C=\frac{a^2+b^2-c^2}{2ab}=\infty$
$\therefore a^2+b^2=c^2$
Using sine rule, we have
$\Rightarrow {\left(2R\sin A\right)}^2+{\left(2R\sin B\right)}^2={\left(2R\sin C\right)}^2$
$\Rightarrow 4R^2{\sin }^{2}A+4R^2{\sin }^{2}B=4R^2{\sin }^{2}C$
Dividing both sides by $4R^2$ we get
$\Rightarrow {\sin }^{2}A+{\sin }^{2}B={\sin }^{2}C$
Add ${\sin }^{2}C$ to both sides, we get
$\Rightarrow {\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2{\sin }^{2}C$
Since $C=\frac{\pi }{2}$ (from above)
$\Rightarrow {\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2{\sin }^2\frac{\pi }{2}$
$\Rightarrow {\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2$
$\therefore \tan A+\tan B=\frac{c^2}{ab}=\frac{a^2+b^2}{ab}=\frac{a}{b}+\frac{b}{a}$
But, $\tan A\tan B=1=\frac{a}{b}\frac{b}{a}$
$\therefore \tan A=\frac{a}{b},\tan B=\frac{b}{a}$