Question

If \( \cosh x = \csc\theta \), then \( \coth^2\left(\frac{x}{2}\right) = \):

• A. \( \tan^2\left(\frac{\theta}{2}\right) \)
• B. \( \tan^2\left(\frac{\pi}{4} - \frac{\theta}{2}\right) \)
• C. \( \cot^2\left(\frac{\theta}{2}\right) \)
• D. \( \cot^2\left(\frac{\pi}{4} - \frac{\theta}{2}\right) \)

Hint:

When solving for hyperbolic functions, use standard identities to convert between trigonometric and hyperbolic forms.

Answer 1

The Correct Option is D

Step 1: Use the hyperbolic identity and given condition.
We are given \( \cosh x = \csc\theta \). Recall the identity for \( \cosh x \) and \( \coth x \): \[ \cosh x = \frac{e^x + e^{-x}}{2}, \quad \coth x = \frac{\cosh x}{\sinh x} \]
Step 2: Simplify using the trigonometric and hyperbolic relationship.
After applying appropriate identities and simplifying the equation: \[ \coth^2\left(\frac{x}{2}\right) = \cot^2\left(\frac{\pi}{4} - \frac{\theta}{2}\right) \]