<s>[INST] $ABC$ is an acute angle triangle, then point $P((\cos B-\sin A),(\sin B-\cos A))$ lies in which quadrant triangle $ABC$ is an acute angle triangle, then point $P((\cos B-\sin A),(\sin B-\cos A))$ lies in which quadrant $.........$
$\cos B-\sin A= \sin(\frac{\pi}{2}-B)-\sin A = 2\cos(\frac{\pi}{2}+A-B)\sin (\frac{\pi}{2}-B-A)$
and $\sin B-\cos A = \cos(\frac{\pi}{2}-B)-\cos A=-2\sin(\frac{\pi}{2}+A-B)\sin (\frac{\pi}{2}-B-A)$
wan,t be able to go further, could some help me with this, thanks  
 [/INST] 
$\sin(\frac{\pi}{2}-B)-\sin A = 2\cos(\frac{\pi}{2}+A-B)\sin (\frac{\pi}{2}-B-A)$
$\cos(\frac{\pi}{2}-B)-\cos A=-2\sin(\frac{\pi}{2}+A-B)\sin (\frac{\pi}{2}-B-A)$

These are incorrect. They should be
$$\sin\left(\frac{\pi}{2}-B\right)-\sin A=2\cos\left(\frac{\frac{\pi}{2}+A-B}{\color{red}{2}}\right)\sin\left(\frac{\frac{\pi}{2}-A-B}{\color{red}{2}}\right)$$
$$\cos\left(\frac{\pi}{2}-B\right)-\cos A=-2\sin\left(\frac{\frac{\pi}{2}+A-B}{\color{red}2}\right)\sin\left(\frac{\frac{\pi}{2}-A-B}{\color{red}2 }\right)$$

Since we have that
$$0\lt A\lt \frac{\pi}{2},-\frac{\pi}{2}\lt -B\lt 0\implies 0\lt \frac{\frac{\pi}{2}+A-B}{2}\lt \frac{\pi}{2}$$
and that
$$0\lt (C=)\pi-A-B\lt\frac{\pi}{2}\implies -\frac{\pi}{4}\lt\frac{\frac{\pi}{2}-A-B}{2}\lt 0$$
we get
$$\cos B-\sin A\lt 0\quad\text{and}\quad \sin B-\cos A\gt 0$$
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