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
${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2$

Answer 1

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

$\tan A,\tan B$

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

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

$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

$=\frac{\frac{c^2}{ab}}{1-1}$

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

$c^2=-a^2+b^2\Rightarrow c^4=a^4+b^4+2a^2{b}^2$

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

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

$\sin A=\frac{b^2}{\sqrt{b^2+a^2}}$

$\sin B=\frac{a^2}{a^2+b^2}$

${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=\frac{b^2+a^2}{b^2+a^2}+1=1+1=2$