Question

If \(\csc \theta + \cot \theta = k\), then the value of \(\csc \theta\) is:

• A. \(\frac{k^2 + 1}{2k}\)
• B. 0
• C. \(\frac{k^2 - 1}{k^2 + 1}\)
• D. \(\frac{1}{k}\)

Hint:

Use the identity \(\csc^2 \theta - \cot^2 \theta = 1\) when working with trigonometric expressions involving \(\csc \theta\) and \(\cot \theta\).

Answer 1

The Correct Option is A

We are given that \(\csc \theta + \cot \theta = k\). Using the identity \(\csc^2 \theta - \cot^2 \theta = 1\), we can square both sides of the given equation: \[ (\csc \theta + \cot \theta)^2 = k^2 \] Expanding: \[ \csc^2 \theta + 2 \csc \theta \cot \theta + \cot^2 \theta = k^2 \] Using the identity \(\csc^2 \theta - \cot^2 \theta = 1\), we substitute: \[ 1 + 2 \csc \theta \cot \theta = k^2 \] Now, solving for \(\csc \theta\) gives: \[ \csc \theta = \frac{k^2 + 1}{2k} \] Thus, the correct answer is option (1).