Question

If $\tan \theta + \sec \theta = m$, then prove that $\sec \theta = \frac{m^2 + 1}{2m}$.

Answer 1

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 \( \sec \theta = \frac{m^2 + 1}{2m} \).