If $A,B \in \left[ 0,\frac{\pi}{2}\right]$ and $\sin{A}\gt\cos{B}$, show that $A+B\gt\frac{\pi}{2}$. The question was this:

If $A,B \in \left[ 0,\frac{\pi}{2}\right]$ and $\sin{A}\gt\cos{B}$, show that $A+B\gt\frac{\pi}{2}$.

Under the topic of trigonometry - Radian & quadrants. And this is not from a textbook. This is from a question paper.
I tried, but I have no idea on how to do this. Please someone help me out :(

First observation is that the $sin$ is monotonically increasing in $[0,\pi/2]$.
Therefore by our hypothesis and considering the addition
formula of sines we get
$$ \sin A > \cos B = \cos B \sin\left(\frac{\pi}{2}\right) - \sin B \cos \left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2} - B\right)$$
Since $\sin$ is monotonic (note that $\pi/2 - B \in [0,\pi/2]$) we get $A > \pi/2 - B$ as expected.
