ΔABC with all acute angles is given. The point whose coordinates are (\(\cos B\;-\;\sin A, \sin B\;-\;\cos A\)) can be in which quadrant.

\( \angle A\;+\;\angle B\;+\;\angle C = \pi\)
\(\Rightarrow \angle A\;+\;\angle B = \pi\; – \;\angle C\)

And from question \(\angle A\; + \; \angle B > \frac{\pi}{2}\)
\(\Rightarrow \angle A > \frac{\pi}{2} – \angle B \)

Taking \(\cos\) on both sides
\(\cos (\frac{\pi}{2} – B) > \cos A\)
\(\Rightarrow \sin B – \cos A > 0\)

Now taking \(\sin\) on both sides
\(\sin A > \sin(\frac{\pi}{2} – B)\)
\(\Rightarrow \sin A > \cos B\)
\(\Rightarrow \cos B – \sin A < 0\)

Therefore Point(\(\cos B\;-\;\sin A, \sin B\;-\;\cos A\)) lies in 2nd quadrant.