We have,
$$
\begin{array}{rrrl}
& & c^2 x^2-c(a+b) x+a b=0 \\
\Rightarrow & c^2 x^2-c a x-c b x+a b=0 \\
\Rightarrow & c x(c x-a)-b(c x-a)=0 \\
\Rightarrow & & (c x-a)(c x-b)=0 \\
\Rightarrow & & c x-a=0 \text { or } c x-b=0 \\
\Rightarrow & x=\frac{a}{c} \text { and } x=\frac{b}{c}
\end{array}
$$
As, $\sin A$ and $\sin B$ are roots of the given equation.
Let $\sin A=\frac{a}{c}$ and $\sin B=\frac{b}{c}$
$\Rightarrow \quad c=\frac{a}{\sin A}$ and $c=\frac{b}{\sin B}$
$\Rightarrow \quad \frac{a}{\sin A}=\frac{b}{\sin B}=c$
Using sine law, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$
So, $\quad \frac{c}{\sin c}=c \quad$ [from Eqs. (i) and (ii)]
$$
\begin{aligned}
& \Rightarrow \quad \sin c=\frac{c}{c} \\
& \Rightarrow \quad \sin C=1 \quad\left[\because \sin \frac{\pi}{2}=1\right] \\
& \Rightarrow \quad \angle C=\frac{\pi}{2} \\
&
\end{aligned}
$$
Since, $\triangle A B C$ is right angled at $\angle c$.

$$
\text { In } \triangle A B C \text {, }
$$




$$
\begin{aligned}
\sin A & =\frac{a}{c} \text { and } \cos A=\frac{b}{c} \\
\therefore \quad \sin A+\cos A & =\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}
\end{aligned}
$$