Solution:In each of the quadrants we draw the graphs

In Quadrant I we can see that
$\sin x\cos x>\tan x$ equals
(decimal (+) less than 1) × (decimal (+) less 1) < (decimal (+)),
this way, the product $\sin x \cdot \cos x$ is (+) but it's less than $\tan x$,
which is also (+) due to the positive decimals comparison criterion.
Conclusion: $\sin x \cdot \cos x>\tan x$ is not met.
In Quadrant II we can see that $\sin x\cos x>\tan x$ equals
(decimal (+) less than 1) × (decimal (-) less than 1) > (decimal (-)),
so the product $\sin x\cos x$ is (-) but greater than $\tan x$,
which is also (-) due to the negative numbers comparison criterion.
Conclusion: $\sin x\cos x>\tan x$ is met.

Quadrant III we can see that
$\sin x\cos x>\tan x$ equals
(decimal (-) less than 0) × (decimal (+) less than 1) < (decimal
(+)),
this way, the product $\sin x\cos x$ is (-) and $\tan x$ is (+).
Conclusion: $\sin x\cos x>\tan x$ is not met.
In Quadrant IV we can see that
$\sin x\cos x>\tan x$ equals
(decimal (-) less than 0) × (decimal (+) less than 1) > (decimal
(-)),
which means the product $\sin x\cos x$ is (-) but greater than $\tan x$
which is also (-) due to the negative numbers comparison criterion.
Conclusion: $\sin x\cos x>\tan x$ is met.
The answer is:
The inequality $\sin x\cos x>\tan x$ is met in quadrants
II and IV.