First of all, I upvoted your query re very positive approach, re
"but I'm looking for a physical derivation for this".
To understand this, in the realm of trigonometry,
where sine and cosine are functions of angles,
you need to consider the sine and cosine functions
against the backdrop of the unit circle.
Imagine a unit circle centered at the origin, that hits the $x$ and $y$
axes at points (1,0), (0,1), (-1,0), and (0, -1).
Consider any point in the unit circle that is in the first (upper right)
quadrant. The point will have coordinates $(x,y).$
Let $\theta$ denote the angle formed by (0,0) -- (1,0) with (0,0) -- (x,y).
Since the radius of the circle is 1, $\cos \theta = x$ and
$\sin \theta = y.$
Now imagine traveling around the arc of the unit circle until you reach the
point (0,1).
This point may be construed to represent $90^{\circ}$, just as one complete
revolution around the circle can be construed to represent $360^{\circ}.$
It is easy to see that $\cos(90^{\circ}) = 0$ and $\sin(90^{\circ}) = 1.$
Now imagine traveling around the arc to any point on the unit circle that
is in the 2nd (upper left) quadrant.
Here, the point $(x,y)$ in the 2nd quadrant will have $x < 0$ and $y > 0.$
Again, just as before, consider $\theta$ to be the angle formed
by (0,0) -- (1,0) with (0,0) -- (x,y).
Here, by convention, $\cos \theta$ (again) $ = x$
and $\sin \theta$ (again) $ = y.$
Thus, it is easy to see that when $(x,y)$ is in the 2nd quadrant, and
$\theta$ is the angle formed by (0,0) -- (1,0) with (0,0) -- (x,y)
that $\cos \theta$ will by convention be $< 0$ and
$\sin \theta$ will by convention be $ > 0.$
A clear advantage of these conventions is that they facilitate the formulas
shown at https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Angle_sum_and_difference_identities.
Another advantage of these conventions, which may not seem important in
the realm of trigonometry, is that they facilitate the cosine and sine functions being continuous functions. This is a pandora's box that may not be worth exploring in the realm of trigonometry, but is still worth a very casual mention.
See https://www.mathopenref.com/triggraphsine.html.
Best Answer
For a less rigorous solution (presumably more suited for the SAT), recall that $\tan(\theta)$ is the slope of the line with angle $\theta$ in the unit circle and $\cos(\theta)$ is the x-coordinate of where that line intersects the unit circle.
In the first quadrant, $\tan(\theta)$ would be greater than $1$ if $\theta>45^{\circ}$. But we also know that the maximum value that $\cos(\theta)$ obtains is $1$, which happens at $\theta=0^{\circ}$. Putting these two facts together gives us (B).