Question

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

Answer 1

Step 1: Given Equation
The provided equation is: \[ \tan \theta + \sec \theta = m \] The objective is to prove: \[ \sec \theta = \frac{m^2 + 1}{2m} \]

Step 2: Express $\tan \theta$ in terms of $\sec \theta$
Using the identity $\tan^2 \theta + 1 = \sec^2 \theta$, we get: \[ \tan \theta = \sqrt{\sec^2 \theta - 1} \] Substituting this into the given equation: \[ \sqrt{\sec^2 \theta - 1} + \sec \theta = m \]

Step 3: Solve for $\sec \theta$
Isolate the square root term: \[ \sqrt{\sec^2 \theta - 1} = m - \sec \theta \] Square both sides: \[ \sec^2 \theta - 1 = (m - \sec \theta)^2 \] Expand and simplify: \[ \sec^2 \theta - 1 = m^2 - 2m \sec \theta + \sec^2 \theta \] \[ -1 = m^2 - 2m \sec \theta \] Rearrange to isolate $\sec \theta$: \[ 2m \sec \theta = m^2 + 1 \] Divide by $2m$: \[ \sec \theta = \frac{m^2 + 1}{2m} \]

Conclusion:
The proof is complete, showing that: \[ \sec \theta = \frac{m^2 + 1}{2m} \]