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

Step 1: Recall the definitions.
If $u = \log_e k$, then $e^u = k$ and $e^{-u} = \tfrac{1}{k}$. Also $\sinh u = \dfrac{e^u - e^{-u}}{2}$.
Step 2: Identify k.
Here $k = \tan\!\left(\tfrac{\pi}{4}+\tfrac{\theta}{2}\right) = \dfrac{1+\tan(\theta/2)}{1-\tan(\theta/2)}$.
Step 3: Write both exponentials.
So $e^u = \dfrac{1+\tan(\theta/2)}{1-\tan(\theta/2)}$ and $e^{-u} = \dfrac{1-\tan(\theta/2)}{1+\tan(\theta/2)}$.
Step 4: Subtract them.
Using a common denominator and $(1+t)^2-(1-t)^2 = 4t$, the difference is $\dfrac{4\tan(\theta/2)}{1-\tan^2(\theta/2)}$.
Step 5: Halve to get sinh.
Then $\sinh u = \dfrac{2\tan(\theta/2)}{1-\tan^2(\theta/2)}$.
Step 6: Recognise the double angle.
This is exactly $\tan\!\left(2\cdot\tfrac{\theta}{2}\right) = \tan\theta$. \[ \boxed{\tan\theta} \]