Step 1: Initial Equation and Objective:
Given the equation \( \tan \theta + \sec \theta = m \), the objective is to prove that \( \sec \theta = \frac{m^2 + 1}{2m} \).
Step 2: Algebraic Manipulation:
Utilize the identity \( \tan^2 \theta + 1 = \sec^2 \theta \) to express \( \tan \theta \) as \( \sqrt{\sec^2 \theta - 1} \). Substitute this into the given equation:
\[
\sqrt{\sec^2 \theta - 1} + \sec \theta = m
\]
Isolate the radical term:
\[
\sqrt{\sec^2 \theta - 1} = m - \sec \theta
\]
Square both sides to remove the radical:
\[
\sec^2 \theta - 1 = (m - \sec \theta)^2
\]
Expand the right side:
\[
\sec^2 \theta - 1 = m^2 - 2m \sec \theta + \sec^2 \theta
\]
Cancel \( \sec^2 \theta \) from both sides:
\[
-1 = m^2 - 2m \sec \theta
\]
Rearrange to solve for \( \sec \theta \):
\[
2m \sec \theta = m^2 + 1
\]
Divide by \( 2m \):
\[
\sec \theta = \frac{m^2 + 1}{2m}
\]
Conclusion:
The derivation confirms that \( \sec \theta = \frac{m^2 + 1}{2m} \).