<s>[INST] Solve $\tan x =\sec 42^\circ +\sqrt{3}$ For the trigonometric equation,
$$\tan x =\sec 42^\circ+\sqrt{3}$$
Find the angle $x$, where $0<x<180^\circ$.
I tried to solve for an unknown angle $x$ in a geometry problem with a trigonometric approach. I ended up with the trig equation above. Without hesitation, I reached my calculator, entering the right-hand-side and arctan-ing it for $x$.
To my surprise, the angle $x$ comes out at exactly 72 degrees. I did not expect such a neat relationship. Then, I thought I should have solved the equation analytically for the whole-degree angle without the calculator. I spent a good amount of time already and  was not able to derive it yet.
Either the equation is not as innocent as it looks, or a straightforward method just eludes me.
 [/INST] Sorry, but I am unable to work with degrees.
If you look here
$$\sec \left(\frac{7 \pi }{30}\right)=\sqrt{8+2 \sqrt{5}-2 \sqrt{15+6 \sqrt{5}}}$$ and here
$$\tan \left(\frac{2 \pi }{5}\right)=\sqrt{5+2 \sqrt{5}}$$
Simplify
$$\left(\sqrt{8+2 \sqrt{5}-2 \sqrt{15+6 \sqrt{5}}}+\sqrt 3\right)^2=5+2 \sqrt{5}$$
I understand your surprise.
Edit
Thinking that this could not be the only one, I computed
$$R_k=\tan \left(\frac{(k+5) \pi}{30}  \right)-\sec \left(\frac{k\pi  }{30}\right)$$ for $k=1,\cdots,60$.
Here are the "funny" results (I hope I did not miss any)
$$\left(
\begin{array}{cc}
k & R_k \\
 5 & \frac{1}{\sqrt{3}} \\
 7 & \sqrt{3} \\
 19 & \sqrt{3} \\
 20 & 2-\frac{1}{\sqrt{3}} \\
 25 & \frac{2}{\sqrt{3}} \\
 30 & 1+\frac{1}{\sqrt{3}} \\
 31 & \sqrt{3} \\
 35 & \frac{5}{\sqrt{3}} \\
 43 & \sqrt{3} \\
 50 & -2-\frac{1}{\sqrt{3}} \\
 55 & -\frac{2}{\sqrt{3}} \\
 60 & -1+\frac{1}{\sqrt{3}}
\end{array}
\right)$$
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