Question

If \[ u=\log\tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right), \] then \[ \tanh\frac{u}{2} = \]

• A. \[ \tan\frac{\theta}{2} \]
• B. \[ \cot\frac{\theta}{2} \]
• C. \[ \sec\frac{\theta}{2} \]
• D. \[ \sin\frac{\theta}{2} \]

Hint:

Remember: $\tanh\frac{u}{2}=\frac{e^u-1}{e^u+1}$.

Answer 1

The Correct Option is A

Step 1: Concept
Express the logarithmic quantity in exponential form and use the definition of hyperbolic tangent.

Step 2: Meaning
Given \[ u=\log\tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right), \] so \[ e^u=\tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right). \]

Step 3: Analysis
Using \[ \tanh\frac{u}{2} = \frac{e^u-1}{e^u+1}, \] we get \[ \tanh\frac{u}{2} = \frac{\tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right)-1} {\tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right)+1}. \] Let \[ t=\tan\frac{\theta}{2}. \] Then \[ \tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right) = \frac{1+t}{1-t}. \] Substituting, \[ \tanh\frac{u}{2} = \frac{\frac{1+t}{1-t}-1} {\frac{1+t}{1-t}+1} = t. \]

Step 4: Conclusion
Therefore \[ \tanh\frac{u}{2} = \tan\frac{\theta}{2}. \]

Final Answer: (A)