Chapter 6 - Analytic Trigonometry - Section 6.4 Trigonometric Identities - 6.4 Assess Your Understanding - Page 497: 30

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

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