If the sines of the angles A and B of a triangle ABC satisfy the equation $c^2{x}^2-c\left(a+b\right)x+ab=0$ , then the triangle

sinA and sinB are the roots of  $c^2{x}^2-c\left(a+b\right)x+ab=0$
Then $\sin A+\sin B=\frac{a+b}{c}$ 
And $\sin A\sin B=\frac{ab}{c^2}$ 
 $\Rightarrow \frac{a}{2R}+\frac{b}{2R}=\frac{a+b}{c}$
And $\frac{a}{2R}\times \frac{b}{2R}=\frac{ab}{c^2}$
$\therefore c=2R\Rightarrow 2R\sin C=2R$

$\therefore \sin C=1$

$\therefore \angle C={90}^{\text{O}}$

$A+B={90}^{\text{O}}$

$B={90}^{\text{O}}-A$

$\because \sin A+\sin B=\frac{a+b}{c}$

$\Rightarrow \sin A+\sin \left({90}^o-A\right)=\frac{a+b}{c}\Rightarrow \sin A+\cos A=\frac{a+b}{c}$