Discussion

If \(\sec \theta + \tan \theta = x\), then \(\sec \theta\) is equal to:

Answer: A

We know that \(\sec^2 \theta - \tan^2 \theta = 1\), which can be factored as \((\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1\).

Given \(\sec \theta + \tan \theta = x\), we have \(\sec \theta - \tan \theta = \frac{1}{x}\).

Now we have a system of two linear equations:

1) \(\sec \theta + \tan \theta = x\)

2) \(\sec \theta - \tan \theta = 1/x\)

Adding the two equations gives: \(2\sec \theta = x + \frac{1}{x} = \frac{x^2+1}{x}\).

Therefore, \(\sec \theta = \frac{x^2+1}{2x}\).

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