Question

\(\cot^2\left(\frac{\pi}{4} + \frac{\theta}{2}\right)\) is equal to

• A. \(\frac{1-\sin\theta}{1+\sin\theta}\)
• B. \(\frac{1-\cos\theta}{1+\cos\theta}\)
• C. \(\frac{1+\sin\theta}{1-\sin\theta}\)
• D. \(\frac{2-\sin\theta}{2+\sin\theta}\)

Hint:

Use half-angle identities for expressions like \(\frac{\pi}{4}+\frac{\theta}{2}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We can solve this by expressing \( \cot \) in terms of \( \sin \) and \( \cos \) or by using the tangent addition formula and taking the reciprocal square.
: Key Formula or Approach:
1. \( \cot^2 A = \frac{1 + \cos 2A}{1 - \cos 2A} \).
2. \( \cos(\pi/2 + \theta) = -\sin \theta \).
Step 2: Detailed Explanation:
Let \( A = \pi/4 + \theta/2 \). We need to evaluate \( \cot^2 A \). Using the half-angle identity for cotangent squared: \[ \cot^2 A = \frac{1 + \cos 2A}{1 - \cos 2A} \] Substitute \( 2A = 2(\pi/4 + \theta/2) = \pi/2 + \theta \): \[ \cot^2 (\pi/4 + \theta/2) = \frac{1 + \cos(\pi/2 + \theta)}{1 - \cos(\pi/2 + \theta)} \] Using the reduction formula \( \cos(\pi/2 + \theta) = -\sin \theta \): \[ \cot^2 (\pi/4 + \theta/2) = \frac{1 - \sin \theta}{1 - (-\sin \theta)} \] \[ \cot^2 (\pi/4 + \theta/2) = \frac{1 - \sin \theta}{1 + \sin \theta} \].
Step 3: Final Answer:
The result is \( \frac{1 - \sin \theta}{1 + \sin \theta} \).