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 :(
 A: 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.
A: $$\sin{A} > \cos{B}$$
$$\sin{A} > \sin{(\frac{\pi}{2} - B)}$$
$$\therefore A > \frac{\pi}{2} - B$$
or
$$\therefore A + B > \frac{\pi}{2}$$
This just makes use of simple trigonometric identities:
For the values that hold this inequality true, look at the graph below:

Here's another way to prove this inequality:


See how the red block has disappeared in image 2... This is because the inequality does not hold true in image 2 and therefore function 1 does not work either (it will only hold true as long as the horizontal blue line is below the horizontal green line). Because cos complements sin (putting it another way: on a 90˚ angle scale, it is essentially the inverse of sin), the value for A changes as B changes and vice-versa, with the relationship: $$A > \frac{\pi}{2} - B$$.
