Question

If \( \theta \) is an acute angle, \( \cosh x = K \) and \( \sinh x = \tan \theta \), then \( \sin \theta = \dots \)

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

Hint:

When solving trigonometric and hyperbolic equations, use known identities such as \( \cosh^2 x - \sinh^2 x = 1 \) to simplify the expression.

Answer 1

The Correct Option is C

We are given that: \[ \cosh x = K \quad {and} \quad \sinh x = \tan \theta. \] We know the fundamental identity for hyperbolic functions: \[ \cosh^2 x - \sinh^2 x = 1. \] Substitute \( \cosh x = K \) and \( \sinh x = \tan \theta \) into this identity: \[ K^2 - \tan^2 \theta = 1. \] Rearrange the equation: \[ K^2 - 1 = \tan^2 \theta. \] Taking the square root of both sides: \[ \sqrt{K^2 - 1} = \tan \theta. \] Now, recall that: \[ \sin \theta = \frac{\tan \theta}{\sqrt{1 + \tan^2 \theta}}. \] Substitute \( \tan \theta = \sqrt{K^2 - 1} \) into the above expression: \[ \sin \theta = \frac{\sqrt{K^2 - 1}}{\sqrt{1 + (K^2 - 1)}} = \frac{\sqrt{K^2 - 1}}{K}. \] Thus, the correct answer is: \[ \boxed{\frac{\sqrt{K^2 - 1}}{K}}. \]