Evaluate trig functions without a calculator 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?
 A: 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.
A: [All angles are measured in degrees]
The (3 – 4 – 5) right-angled triangle gives the closest approximation of angle equal to 53 degrees (53.1xxxxx degrees to be exact, slightly larger than the required 53 degree angle).
Thus, we construct a right angled triangle of sides (3 + 4d, 4 – 3d, 5); where d is a very small positive increment in the range 50d (at the most) is equal to 1). [The reason of using 4d and 3d will be clear when simplification of the following is performed.]
Then, $\tan 52 = \dfrac {4 – 3d}{3 + 4d}$ and $\sec 38 = \dfrac {5}{4 – 3d}$
$\tan 52 – \sec 38 = …. = \dfrac {1 – 49d + 9d^2}{(3+4d)(4 – 3d)} = ... > 0$ (since d is small as assumed above)
Therefore, $\tan 52 > \sec 38$
Since tan is an increasing function in the range $[0, 90), therefore, \tan 53 > \sec 38$
Remark: Although $(3 + 4d, 4 – 3d, 5)$ is not exactly right-angled, $(3 + 4d, 4 – 3d, \sqrt(25 + 25d^2))$ is. Since d is small enough, we can use that as a close estimate. 
A: Here's one way to do this without protractors and such, with some algebra and basic trigonometric identities. First, notice that the sum and difference of the two angles are 91° and 15°, and the values of trigonometric functions at both these angles are either known or very easy to evaluate. This gives one the idea of transforming the inequality $\tan53° ≶ \sec38°$ in such a way that the awkward angles disappear and the sums and differences of angles come out instead. Since $\tan53°$ and $\sec38°$ are both positive, their squares are related in the same way as the quantities themselves:$$\tan^2 53° ≶  \sec^2 38°,$$or$$\tan^2 53° - \sec^2 38° ≶ 0.$$Transforming the left-hand side with the goal of obtaining sums and differences of the angles,$$LHS=\tan^2 53°-\frac{1}{\cos^2 38°}=\tan^2 53°-(\tan^2 38°+1)=\frac{\sin^2 53°}{\cos^2 53°}-\frac{\sin^2 38°}{\cos^2 38°}-1=$$$$=\frac{\sin^2 53°\cos^2 38°-\sin^2 38°\cos^2 53°}{\cos^2 53°\cos^2 38°}-1=\frac{\sin(53°-38°)\sin(53°+38°)}{((\cos(53°+38°)+\cos(53°-38°))/2)^2}-1=$$ $$=\frac{4\sin15°\sin91°}{(\cos15°+\cos91°)^2}-1$$
This expression is much easier to evaluate, because $\sin91°$ is very close to $1$ and $\cos91°$ is rather close to $0$ (less than 0.02 by absolute value). Neglecting these small quantities for the moment, we find that$$LHS\approx\frac{4\sin15°}{\cos^2 15°}-1=\frac{4\sin15°\cos15°}{\cos^3 15°}-1=\frac{2\sin 30°}{\cos^3 15°}-1=\frac{1}{\cos^3 15°}-1.$$Since the cosine is always smaller than $1$ this quantity is obviously positive, and 15° is a large enough angle that the small quantities we neglected above will not be able to affect the sign of $LHS$. Hence, the correct sign in the original inequality is $>$:$$\tan53°>\sec38°.$$
A: 
I think what this all boils down to is that we have to show that 
$$\cos 54^\circ<\cos 53^\circ<1-\phi\approx \cos 51.8\dots^\circ$$
If we calculate the value of the derivative at $54^\circ$ of the cosine function (differentiating with respect of radians of course) and draw a line whose slope is that value, passing through $(54^\circ,\cos 54^\circ)$, we can use that line to estimate the value of $\cos 53^\circ$. Because of the concavity of the cosine function, we know that the the estimate is greater than the actual value of $\cos 53^\circ$. Thus, if we can show that the estimate is less than $1-\phi$, then the problem is solved.
The actual value of the arccosine of the estimate is approx. $52.99\dots^\circ$
