Question

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

Answer 1

Step 1: Start with the given equation:
We are given that: \[ \tan \theta + \sec \theta = m \] We need to prove that: \[ \sec \theta = \frac{m^2 + 1}{2m} \]

Step 2: Express $\tan \theta$ in terms of $\sec \theta$:
We know that: \[ \tan^2 \theta + 1 = \sec^2 \theta \] So, we can express $\tan \theta$ as: \[ \tan \theta = \sqrt{\sec^2 \theta - 1} \] Substitute this into the given equation $\tan \theta + \sec \theta = m$: \[ \sqrt{\sec^2 \theta - 1} + \sec \theta = m \] This is the key equation we will work with.

Step 3: Isolate the square root term:
Move $\sec \theta$ to the other side: \[ \sqrt{\sec^2 \theta - 1} = m - \sec \theta \] Now, square both sides of the equation to eliminate the square root: \[ \sec^2 \theta - 1 = (m - \sec \theta)^2 \] Expand the right-hand side: \[ \sec^2 \theta - 1 = m^2 - 2m \sec \theta + \sec^2 \theta \] Simplify by canceling the $\sec^2 \theta$ terms on both sides: \[ -1 = m^2 - 2m \sec \theta \] Rearrange the terms to isolate $\sec \theta$: \[ 2m \sec \theta = m^2 + 1 \] Finally, divide both sides by $2m$ to solve for $\sec \theta$: \[ \sec \theta = \frac{m^2 + 1}{2m} \]

Conclusion:
We have proven that: \[ \sec \theta = \frac{m^2 + 1}{2m} \]