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Question

If $P = \tan\theta + \sec\theta$, then the value of $\tan\theta$ is:

This question was previously asked in
RRB NTPC 2024 CBT 1 Question Paper (28-Aug-2025) (Shift 3)
The correct answer is
$\frac{P^2 - 1}{2P}$

Finding tan theta Value Using P = tan theta + sec theta

We are given the expression $P = \tan\theta + \sec\theta$. Our goal is to find the value of $\tan\theta$ in terms of $P$.

Using Trigonometric Identities

We know the fundamental trigonometric identity: $ \sec^2\theta - \tan^2\theta = 1 $ This can be factored as the difference of squares:

$ (\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1 $

Substitute the given expression $P = \tan\theta + \sec\theta$ into the factored identity:

$ (\sec\theta - \tan\theta) P = 1 $

Now, we can express $(\sec\theta - \tan\theta)$ in terms of $P$:

$ \sec\theta - \tan\theta = \frac{1}{P} $

Solving for tan theta

We now have a system of two linear equations involving $\sec\theta$ and $\tan\theta$:

  1. $ \sec\theta + \tan\theta = P $
  2. $ \sec\theta - \tan\theta = \frac{1}{P} $

To find $\tan\theta$, subtract the second equation from the first:

$ (\sec\theta + \tan\theta) - (\sec\theta - \tan\theta) = P - \frac{1}{P} $

Simplify the left side:

$ \sec\theta + \tan\theta - \sec\theta + \tan\theta = P - \frac{1}{P} $

$ 2\tan\theta = P - \frac{1}{P} $

Combine the terms on the right side using a common denominator:

$ 2\tan\theta = \frac{P \cdot P}{P} - \frac{1}{P} $

$ 2\tan\theta = \frac{P^2 - 1}{P} $

Finally, divide by 2 to isolate $\tan\theta$:

$ \tan\theta = \frac{P^2 - 1}{2P} $

This matches the value provided in Option D.

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