Question

If sec θ+ tan θ= X, then tan θ =

• A. \(\frac{x^2+1}{x}\)
• B. \(\frac{x^2-1}{x}\)
• C. \(\frac{x^2+1}{2x}\)
• D. \(\frac{x^2-1}{2x}\)

Answer 1

The Correct Option is D

We are given that: \[ \sec \theta + \tan \theta = x \] We need to find \( \tan \theta \) in terms of \( x \). First, recall the identity: \[ \sec^2 \theta - \tan^2 \theta = 1 \] Now, square both sides of the given equation \( \sec \theta + \tan \theta = x \): \[ (\sec \theta + \tan \theta)^2 = x^2 \] Expanding the left-hand side: \[ \sec^2 \theta + 2 \sec \theta \tan \theta + \tan^2 \theta = x^2 \] Using the identity \( \sec^2 \theta - \tan^2 \theta = 1 \), substitute this into the equation: \[ 1 + 2 \sec \theta \tan \theta = x^2 \] Rearranging: \[ 2 \sec \theta \tan \theta = x^2 - 1 \] Now, solve for \( \sec \theta \tan \theta \): \[ \sec \theta \tan \theta = \frac{x^2 - 1}{2} \] We know that \( \sec \theta = \frac{1}{\cos \theta} \), and \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Therefore: \[ \frac{1}{\cos \theta} \cdot \frac{\sin \theta}{\cos \theta} = \frac{x^2 - 1}{2} \] \[ \frac{\sin \theta}{\cos^2 \theta} = \frac{x^2 - 1}{2} \] Finally, solving for \( \tan \theta \): \[ \tan \theta = \frac{x^2 - 1}{2x} \]

The correct option is (D): \(\frac{x^2-1}{2x}\)