In a right angled triangle ABC,

Let, $\angle B=90^{\circ}$

$\therefore A+C=90^{\circ}=\frac{\pi}{2}$

$\Rightarrow C=\frac{\pi}{2}-A$

$\Rightarrow \sin C=\sin \left(\frac{\pi}{2}-A\right)$

$\Rightarrow \sin C=\cos A \ldots(\mathrm{i})$

Now,

$\sin ^{2} A+\sin ^{2} B+\sin ^{2} C=\sin ^{2} A+1+\sin ^{2} C \quad\left(\because \sin B=\sin 90^{\circ}=1\right)$

$=\sin ^{2} A+\cos ^{2} A+1 \quad[$ Using $(\mathrm{i})]$

$=1+1$

$=2$