Chapter 7 - 7.1 - Using Fundamental Identities - 7.1 Exercises - Page 514: 72

$\sqrt {a^2+u^2}=a~sec~\theta$

$u=a~tan~\theta$ $\sqrt {a^2+u^2}=\sqrt {a^2+a^2tan^2\theta}=\sqrt {a^2(1+tan^2\theta)}=\sqrt {a^2sec^2\theta}=\pm~a~sec~\theta$ Given that $-\frac{\pi}{2}\lt\theta\lt\frac{\pi}{2}$ $\sqrt {a^2+u^2}=a~sec~\theta$