If $$\frac{1 + sin \phi}{1 - sin \phi} = \frac{p^2}{q^2}$$, then $$sec \phi$$ is equal to 

$$\frac{1 + sin \phi}{1 - sin \phi} = \frac{p^2}{q^2}$$
By componendo dividendo rule,
$$\frac{(1 + sin \phi) + (1 - sin \phi)}{(1 + sin \phi) - (1 - sin \phi)} = \frac{p^2 + q^2}{p^2 - q^2}$$
$$\frac{2}{2 sin \phi} = \frac{p^2 + q^2}{p^2 - q^2}$$
$$ sin \phi = \frac{p^2 - q^2}{p^2 + q^2}$$ = $$\frac{perpendicular}{hypotenuse}$$
By the Pythagoras theorem,
Base = $$\sqrt{(p^2 + q^2)^2- (p^2 - q^2)^2 } = \sqrt{4p^2q^2} = 2pq$$
$$sec \phi$$ =  $$\frac{hypotenuse}{base} = \frac{p^2 + q^2}{2pq}$$