Determine if the following is an identity: $$\displaystyle{\sec^2 \theta + \csc^2 \theta = \sec^2 \theta \csc^2 \theta}$$

First we attempt to simplify the left side:

$$\begin{array}{rclc} \sec^2 \theta + \csc^2 \theta &=& \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta} & \scriptsize{\textrm{rewriting in terms of } \sin \theta \textrm{ and } \cos \theta}\\\\ &=& \frac{\sin^2 \theta}{\cos^2 \theta \sin^2 \theta} + \frac{\cos^2 \theta}{\cos^2 \theta \sin^2 \theta} & \overset{\normalsize{\textrm{getting common denominators}}}{\scriptsize{\textrm{so these fractions can be combined}}}\\\\ &=& \frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta \sin^2 \theta} & \scriptsize{\textrm{now recall } \sin^2 \theta + \cos^2 \theta = 1}\\\\ &=& \frac{1}{\cos^2 \theta \sin^2 \theta} & \end{array}$$

Which gets us to within one step of the right side, as:

$$\begin{array}{rclc} \sec^2 \theta \csc^2 \theta &=& \frac{1}{\cos^2 \theta \sin^2 \theta} & \quad \scriptsize{\textrm{rewriting in terms of } \sin \theta \textrm{ and } \cos \theta}\\\\ \end{array}$$

Thus, the equation above is an identity.