`tantheta+(1)/(3)tan^(3)theta+(1)/(5)tan^(5)theta+....=`

Options

A. `(1)/(2)log[sin((pi)/(4)+theta)]`
B. `(1)/(2)log[cos((pi)/(4)+theta)]`
C. `(1)/(2)log[tan((pi)/(4)+theta)]`
D. `(1)/(2)log[cot((pi)/(4)+theta)]`

Answer

To solve the given series \( \tan \theta + \frac{1}{3} \tan^3 \theta + \frac{1}{5} \tan^5 \theta + \ldots \), we can recognize that this series resembles the expansion of the logarithmic function.

### Step-by-Step Solution:

1. **Identify the Series**:
The series can be expressed as:
\[ S = \tan \theta + \frac{1}{3} \tan^3 \theta + \frac{1}{5} \tan^5 \theta + \ldots \]

2. **Recall the Logarithmic Series**:
We know from logarithmic identities that:
\[ \log \left( \frac{1+x}{1-x} \right) = 2x + \frac{x^3}{3} + \frac{x^5}{5} + \ldots \]
This is valid for \( |x| < 1 \).

3. **Substitute \( x = \tan \theta \)**:
By substituting \( x = \tan \theta \) into the logarithmic series, we have:
\[ \log \left( \frac{1+\tan \theta}{1-\tan \theta} \right) = 2 \tan \theta + \frac{\tan^3 \theta}{3} + \frac{\tan^5 \theta}{5} + \ldots \]

4. **Equate the Series**:
From our series \( S \), we can equate:
\[ S = \tan \theta + \frac{1}{3} \tan^3 \theta + \frac{1}{5} \tan^5 \theta + \ldots = \frac{1}{2} \log \left( \frac{1+\tan \theta}{1-\tan \theta} \right) \]

5. **Use the Identity for \( \tan \frac{\pi}{4} \)**:
We know that \( \tan \frac{\pi}{4} = 1 \). Therefore, we can express:
\[ \frac{1+\tan \theta}{1-\tan \theta} = \frac{1+\tan \theta}{1-\tan \theta} = \tan \left( \frac{\pi}{4} + \theta \right) \]

6. **Final Expression**:
Thus, we can write:
\[ S = \frac{1}{2} \log \left( \tan \left( \frac{\pi}{4} + \theta \right) \right) \]

### Conclusion:
The final result is:
\[ S = \frac{1}{2} \log \left( \tan \left( \frac{\pi}{4} + \theta \right) \right) \]