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Question

If $\sec A + \tan A = p$, then what is $\sec A$ in terms of p?

This question was previously asked in
SSC CGL 2025 Tier 1 Question Paper (25-Sep-2025) (Shift 3)
The correct answer is
$\frac{p^2 + 1}{2p}$

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:

  1. $ \sec A + \tan A = p $
  2. $ \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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