So, the best way to do this is to go back to basics of a right triangle!!
|dw:1321912053720:dw|
H = hypotenuse, B = base, P = perpendicular
\[\sec(x) = \frac{H}{B}\]
\[\sec^2(x) = \frac{H^2}{B^2}\]
\[Sec^2(x) - 1 = \frac{H^2}{B^2} - 1 = \frac{H^2 - B^2}{B^2}\]
Now, according to the Pythagoras theorem,
\[H^2 = P^2 + B^2\]
Which means
\[H^2 - B^2 = P^2\]
So,
\[Sec^2(x) - 1 = \frac{H^2 - B^2}{B^2} = \frac{P^2}{B^2}\]
\[\frac{P}{B} = tan(x)\]
Thus, you get your answer :)