Establish the identity. \( \begin{array}{l}1+\frac{\tan ^{2} \theta}{\sec \theta+1}=\sec \theta \\ \text { Simplify the numerator of the fraction from the left side by applying the appropriate Pythagorean Identity. } \\ 1+\frac{\sec ^{2} \theta-1}{\sec \theta+1} \text { (Do not factor.) } \\ \text { Factor the numerator of the fraction from the previous step. } \\ 1+\frac{\square}{\sec \theta+1}\end{array} \)

To establish the identity \[ 1+\frac{\tan ^{2} \theta}{\sec \theta+1}=\sec \theta, \] we will follow the steps outlined in the problem. ### Step 1: Simplify the numerator of the fraction using the Pythagorean Identity We know from the Pythagorean identity that \[ \tan^2 \theta + 1 = \sec^2 \theta. \] Thus, we can express \(\tan^2 \theta\) as: \[ \tan^2 \theta = \sec^2 \theta - 1. \] Substituting this into the left side of the identity gives us: \[ 1 + \frac{\tan^2 \theta}{\sec \theta + 1} = 1 + \frac{\sec^2 \theta - 1}{\sec \theta + 1}. \] ### Step 2: Factor the numerator of the fraction Now we need to simplify the fraction: \[ 1 + \frac{\sec^2 \theta - 1}{\sec \theta + 1}. \] The numerator \(\sec^2 \theta - 1\) can be factored as: \[ \sec^2 \theta - 1 = (\sec \theta - 1)(\sec \theta + 1). \] Now substituting this back into the expression gives us: \[ 1 + \frac{(\sec \theta - 1)(\sec \theta + 1)}{\sec \theta + 1}. \] ### Step 3: Simplify the expression Since \(\sec \theta + 1\) is in both the numerator and the denominator, we can cancel it out (assuming \(\sec \theta + 1 \neq 0\)): \[ 1 + (\sec \theta - 1). \] This simplifies to: \[ \sec \theta. \] ### Conclusion Thus, we have shown that: \[ 1 + \frac{\tan^2 \theta}{\sec \theta + 1} = \sec \theta, \] which establishes the identity.