How does $\tan 70^\circ - \sec 10^\circ$ have the exact value of $\sqrt{3}$? Recently, while trying to solve a problem posed by the Youtube video here, I found the following relation:
$$\tan 70^\circ - \sec 10^\circ = \sqrt{3}$$
This relation is exact, and that can be proved by combining the purely geometric arguments presented in the video linked above and some basic trigonometry. 
But as far as I know, the trigonometric ratios of $10^\circ$ and $70^\circ$ cannot be expressed in exact form (using only square roots). In fact, Gauss proved that the sine or cosine of any angle ${360^\circ}\over n$ cannot be expressed in terms of fractions and square roots unless $n = 2^m\cdot\prod p_i$ where every $p_i$ is a Fermat prime: neither $10^\circ$ nor $70^\circ$ is even constructible.
So how does the stated relation hold? I have been unable to prove why it should ... any insights appreciated! 

Great answer by Michael! Since his answer was a bit terse and I had to spend some time figuring it out, here is an expansion. Since $\tan 60^\circ = \sqrt{3}$, and because $\sec 10^\circ = \frac {1} {\cos 10^\circ}$ by definition,  we can say that we need to prove
$$
\frac {1} {\cos 10^\circ} = \tan 70^\circ - \tan 60^\circ
$$
Since $\sin \theta = ( \cos 90^\circ - \theta )$, and since $\sin 30^\circ = {1 \over 2}$, the RHS further reduces as follows:
$$
 = \frac {\sin 70^\circ} {\cos 70^\circ} - \frac {\sin 60^\circ} {\cos 60^\circ} = \frac {\cos 20^\circ} {\sin 20^\circ} - \frac {\cos 30^\circ} {\sin 30^\circ}
= \frac {\sin 30^\circ \cdot \cos 20^\circ - \cos 30^\circ \cdot \sin 20^\circ} {\sin 20^\circ \cdot \sin 30^\circ}
= \frac {\sin (30^\circ - 20^\circ)} {\sin 20^\circ \cdot \sin 30^\circ}
= \frac {\sin 10^\circ} {\sin 20^\circ \cdot \sin 30^\circ}
=2\cdot\frac{\sin 10^\circ} {\sin 20^\circ} = 2\cdot\frac{\sin 10^\circ} {2\cdot \sin 10^\circ \cdot \cos 10^\circ}
=\frac {1} {\cos 10^\circ}
$$
