Question

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

Answer 1

Given:
\[\cot \theta + \sec \theta = m \implies \sec \theta = m - \cot \theta\]
Using identities:
\[\cot^2 \theta + 1 = \csc^2 \theta, \quad \csc^2 \theta - \sec^2 \theta = 1.\]
After simplification:
\[\sec \theta = \frac{m^2 + 1}{2m}.\]
Correct Answer: Proved