Answer
$\sqrt {x^2-16}=4~tan~θ$
Work Step by Step
$1+tan^2θ=sec^2θ$
$sec^2θ-1=tan^2θ$
$\sqrt {x^2-16}=\sqrt {(4~sec^2θ)^2-16}=\sqrt {16~sec^2θ-16}=\sqrt {16(sec^2θ-1)}=4\sqrt {tan^2~θ}=±4~tan~θ$
But, since $0\lt θ\lt\frac{\pi}{2}$:
$\sqrt {x^2-16}=4~tan~θ$
