Trigonometric Relationship
We are given the equation: $ \sec A + \tan A = p $
We will use the fundamental trigonometric identity: $ \sec^2 A - \tan^2 A = 1 $
Algebraic Manipulation
Factor the identity as a difference of squares: $ (\sec A - \tan A)(\sec A + \tan A) = 1 $
Substitute the given value of $p$: $ (\sec A - \tan A)(p) = 1 $
Solve for $(\sec A - \tan A)$: $ \sec A - \tan A = \frac{1}{p} $
Solving for Sec A
Now we have a system of two linear equations:
- $ \sec A + \tan A = p $
- $ \sec A - \tan A = \frac{1}{p} $
Add the two equations to eliminate $\tan A$: $ (\sec A + \tan A) + (\sec A - \tan A) = p + \frac{1}{p} $ $ 2 \sec A = p + \frac{1}{p} $
Combine the terms on the right side: $ 2 \sec A = \frac{p^2 + 1}{p} $
Isolate $\sec A$ by dividing by 2: $ \sec A = \frac{p^2 + 1}{2p} $
This matches option A.
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