Question

If \( \theta = \sec^{-1}(\cosh u) \), then \( u = \)

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

Hint:

When working with inverse trigonometric and hyperbolic functions, use standard identities to convert them into logarithmic forms and simplify using angle sum or difference identities.

Answer 1

The Correct Option is D

Step 1: Recognize that \(\sec^{-1}(x)\) returns an angle \(\theta\) such that \( \cos \theta = \frac{1}{x} \). In this case, we have: \[ \theta = \sec^{-1}(\cosh u) \implies \cos \theta = \frac{1}{\cosh u} \] Step 2: Relate the inverse secant function to the hyperbolic cosine function. We know that: \[ \sec \theta = \cosh u \implies \cos \theta = \frac{1}{\cosh u} \] Step 3: Since \(\cosh u\) and \(\sec \theta\) are hyperbolic and trigonometric functions respectively, solving for \(u\) involves using their inverse relationships: \[ u = \cosh^{-1}(\sec \theta) \] Step 4: Express the inverse hyperbolic cosine function in logarithmic form using the standard identity: \[ \cosh^{-1}(x) = \log_e(x + \sqrt{x^2 - 1}) \] Substituting \( \sec \theta \) for \( x \): \[ u = \log_e \left( \sec \theta + \sqrt{\sec^2 \theta - 1} \right) \] Step 5: Simplify further using the identity \( \sec^2 \theta - 1 = \tan^2 \theta \), so the expression becomes: \[ u = \log_e \left( \sec \theta + \tan \theta \right) \] Step 6: Now express \( \sec \theta \) and \( \tan \theta \) in terms of \( \theta \). The relation between \( \sec \theta \) and \( \tan \theta \) can be simplified using angle identities: \[ u = \log_e \left( \tan \left( \frac{\pi}{4} - \frac{\theta}{2} \right) \right) \]