Chapter 6 - Analytic Trigonometry - Section 6.4 Trigonometric Identities - 6.4 Assess Your Understanding - Page 497: 29

The left side is equivalent to $1$ therefore the given equation is an identity. Refer to the solution below.

$\text{ LHS } = (\sec{\theta}+\tan{\theta})(\sec{\theta}-\tan{\theta}) $ $\text{By Expanding:}$ \begin{align} \text{ LHS } & = \sec^2{\theta}+\sec{\theta}\tan{\theta}-\sec{\theta}\tan{\theta}- \tan^2{\theta} \\[2mm] & = \sec^2{\theta}-\tan^2{\theta} \\[2mm] &= 1 \\[2mm] & = \text{ RHS} \end{align}