How does $\tan 70^\circ - \sec 10^\circ$ have the exact value of $\sqrt{3}$?

We need to prove that: $$\tan70^{\circ}-\tan60^{\circ}=\frac{1}{\cos10^{\circ}}$$ or $$\frac{\sin10^{\circ}}{\cos70^{\circ}\cos60^{\circ}}=\frac{1}{\cos10^{\circ}}$$ or $$2\sin10^{\circ}\cos10^{\circ}=\sin20^{\circ},$$ which is true.

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The first step follows from the sine addition subtraction rules and the tangent rules:

$$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta\\=\left(\frac{\sin\alpha}{\cos\alpha}\pm\frac{\sin\beta}{\cos\beta}\right)(\cos\alpha\cos\beta)=(\tan\alpha\pm\tan\beta)(\cos\alpha\cos\beta)\\\tan\alpha\pm\tan\beta=\frac{\sin(\alpha\pm\beta)}{\cos\alpha\cos\beta}$$

The second step follows because $\cos60^\circ=\frac12$ and $\cos70^\circ=\sin20^\circ$

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: $$ = \frac {\sin 70^\circ} {\cos 70^\circ} - \frac {\sin 60^\circ} {\cos 60^\circ} = \frac {\cos 20^\circ} {\sin 20^\circ} - \frac {\cos 30^\circ} {\sin 30^\circ} = \frac {\sin 30^\circ \cdot \cos 20^\circ - \cos 30^\circ \cdot \sin 20^\circ} {\sin 20^\circ \cdot \sin 30^\circ} = \frac {\sin (30^\circ - 20^\circ)} {\sin 20^\circ \cdot \sin 30^\circ} = \frac {\sin 10^\circ} {\sin 20^\circ \cdot \sin 30^\circ} =2\cdot\frac{\sin 10^\circ} {\sin 20^\circ} = 2\cdot\frac{\sin 10^\circ} {2\cdot \sin 10^\circ \cdot \cos 10^\circ} =\frac {1} {\cos 10^\circ} $$