In \( \Large \triangle ABC \), if \( \Large \angle C \) is 90 \(^{\circ} \), then

\( \Large \angle A+ \angle B \) = 180  \(^{\circ} \) - 90  \(^{\circ} \) = 90  \(^{\circ} \)

Now, cos (A + B) + sin(A + B)

=cos 90  \(^{\circ} \)+ sin 90  \(^{\circ} \) = 0+1=1