Chapter 7 - 7.1 - Using Fundamental Identities - 7.1 Exercises - Page 514: 55

$\sqrt {x^2-4}=2~tan~θ$

$sec^2θ=1+tan^2θ$ $sec^2θ-1=tan^2θ$ $\sqrt {x^2-4}=\sqrt {(2~sec~θ)^2-4}=\sqrt {4~sec^2θ-4}=\sqrt {4(sec^2θ-1)}=2\sqrt {tan^2θ}=±2~tan~θ$ But, since $0\ltθ\lt\frac{\pi}{2}$ (Quadrant I): $\sqrt {x^2-4}=2~tan~θ$