Question

If \( u=log_{e}[tan(\frac{\pi}{4}+\frac{\theta}{2})], \) then \( sinh~u= \)

• A. \( cos~\theta \)
• B. \( sec~\theta \)
• C. \( tan~\theta \)
• D. \( sin~\theta \)

Hint:

The logarithmic tangent half-angle form is a classic relationship in hyperbolic-trigonometric conversions. Keep these two core transformations handy: \[ sinh(u) = tan(\theta) \] \[ cosh(u) = sec(\theta) \]

Answer 1

The Correct Option is C

Concept: By exponential definitions, if \( u = \log_e(k) \), then \( e^u = k \) and \( e^{-u} = \frac{1}{k} \). The definition of the hyperbolic sine function is: \[ sinh~u = \frac{e^u - e^{-u}}{2} \]

Step 1: Finding expressions for exponential values.
Here, \( k = tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right) = \frac{1+tan(\theta/2)}{1-tan(\theta/2)} \). Thus, we have: \[ e^u = \frac{1+tan(\theta/2)}{1-tan(\theta/2)} \] \[ e^{-u} = \frac{1-tan(\theta/2)}{1+tan(\theta/2)} \]

Step 2: Computing the definition of \( sinh~u \).
\[ sinh~u = \frac{1}{2} \left[ \frac{1+tan(\theta/2)}{1-tan(\theta/2)} - \frac{1-tan(\theta/2)}{1+tan(\theta/2)} \right] \] Taking a common denominator: \[ = \frac{1}{2} \left[ \frac{(1+tan(\theta/2))^2 - (1-tan(\theta/2))^2}{1 - tan^2(\theta/2)} \right] \] Using the algebraic identity \( (1+x)^2 - (1-x)^2 = 4x \): \[ = \frac{1}{2} \left[ \frac{4tan(\theta/2)}{1 - tan^2(\theta/2)} \right] = \frac{2tan(\theta/2)}{1 - tan^2(\theta/2)} \]

Step 3: Converting to double-angle form.
The expression matches the standard double-angle identity for tangent: \[ \frac{2tan(\theta/2)}{1 - tan^2(\theta/2)} = tan\left(2 \times \frac{\theta}{2}\right) = tan~\theta \]