My precalculus test asked me to determine which was greater: $\tan (53)$ or $\sec (38)$.
I looked at it like this, but it seems so close that it's difficult to imagine that they would ask this:
$\tan (45)$ is 1 and $\tan (60)$ is $\sqrt{3}$, so since 53 is approx between 45 and 60, I took a value in-between $1$ and $1.73\ldots$ say $1.36$
$\dfrac{1}{\cos (30)}$ is about $1.2$, and
$\dfrac{1}{\cos (45)}$ is about $1.4.$
Taking a value in-between I chose $1.3$
So, obviously I was correct to choose the tangent value as being larger, but it is actually larger by about $.057$. HOW AM I SUPPOSED TO DO THIS without a calculator?
Best Answer
Not an answer, but a discussion of the closeness of the situation.
Since $53^\circ$ and $38^\circ$ are very nearly complementary, we have that $\sec 38^\circ \approx \csc 53^\circ$ ... with the left-hand side being ever-so-slightly larger than the right-hand side.
As the first diagram suggests, for big enough (first-quadrant) angles $\theta$, we have that $\tan\theta$ exceeds $\csc\theta$; and, according to that first diagram, $53^\circ$ seems to be one of those "big enough" angles ... but just barely. Is it big enough that the $\tan 53^\circ$ also exceeds the slightly-larger value, $\sec 38^\circ$? Well, the middle diagram confirms that it is (though again: just barely), but of course having a computer program draw an accurate diagram is really no better than using a calculator compute the values.
What makes the approximations especially-tricky here is that $53^\circ$ is very close to the magic (or, should I say, "golden"?) angle, $\theta_\star = 51.8...^\circ$, marking the threshold of those "big enough" angles. If the problem had been to compare, say, $\tan 70^\circ$ with $\sec 21^\circ$, then we would have had more confidence in our ability to fiddle with the numbers.
All things considered, this seems like a bad exercise for a test. I wonder if there was an error in the test question.