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.
 A: $\begin{align} \cos(42°) &= \cos(60°-18°) \cr
&= \cos(60°)\cos(18°) + \sin(60°)\sin(18°) \cr
&= {1\over2} (\cos(18°) + \sqrt3 \sin(18°)) \cr
\sec(42°) &= \left({2 \over \cos(18°) + \sqrt3 \sin(18°)}\right) 
\left({\cos(18°) - \sqrt3 \sin(18°) \over \cos(18°) - \sqrt3 \sin(18°)}\right) \cr
&= {2(\cos(18°) - \sqrt3 \sin(18°)) \over \cos^2(18°) - 3\sin^2(18°)} \cr
&= \left({2\sin(18°) \over 1 -4 \sin^2(18°)}\right) (\cot(18°) - \sqrt3) \cr
\end{align}$ 
Let $s=\sin(18°)$, using multiple angles formula
$\sin(90°) = \sin(5 \times 18°) = 16s^5 - 20s^3 + 5s = 1$ 
$16s^5 - 20s^3 + 5s - 1 = 0$
$(s-1)(4s^2+2s-1)^2 = 0$
Since $s≠1$, we get $4s^2+2s-1 = 0\quad → \large{2s \over 1-4s^2} = 1$
$\tan(x) = \sec(42°) + \sqrt3 = (\cot(18°) - \sqrt3) + \sqrt3 = \tan(72°)$
A: 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)
$$\left(
\begin{array}{cc}
k & R_k \\
 5 & \frac{1}{\sqrt{3}} \\
 7 & \sqrt{3} \\
 19 & \sqrt{3} \\
 20 & 2-\frac{1}{\sqrt{3}} \\
 25 & \frac{2}{\sqrt{3}} \\
 30 & 1+\frac{1}{\sqrt{3}} \\
 31 & \sqrt{3} \\
 35 & \frac{5}{\sqrt{3}} \\
 43 & \sqrt{3} \\
 50 & -2-\frac{1}{\sqrt{3}} \\
 55 & -\frac{2}{\sqrt{3}} \\
 60 & -1+\frac{1}{\sqrt{3}}
\end{array}
\right)$$
A: Golden triangles,  I use these mnemonic:
$\;\displaystyle \cos 36° = \frac{\sqrt{2+\frac{1}{ϕ}}}{2} = \frac{ϕ}{2}\;,\;
\sin 36° = \frac{\sqrt{2-\frac{1}{ϕ}}}{2}$
$\;\displaystyle \cos 18° = \frac{\sqrt{2+ϕ}}{2} \qquad\;\;,\;
\sin 18° = \frac{\sqrt{2-ϕ}}{2} = \frac{1}{2ϕ}$

$\displaystyle \cos (60°-18°) = 
\frac{1}{2} × \frac{\sqrt{2+ϕ}}{2} \;+\;
\frac{\sqrt{3}}{2} × \frac{1}{2ϕ}$
$\displaystyle \sec 42° = \frac{4ϕ}{ϕ\sqrt{2+ϕ} \,+ \sqrt{3}}
= \frac{4ϕ\,(ϕ\sqrt{2+ϕ} \,- \sqrt{3})}{ϕ^2\,(2+ϕ)-3}
= ϕ\sqrt{2+ϕ} \,- \sqrt{3}$
First term: $\;ϕ\sqrt{2+ϕ} \,= \cot 18° = \tan 72°$
