If ` ( tan theta - i [ sin ( theta // 2) + cos ( theta // 2) ])/( 1 + 2 i sin ( theta // 2))` is purely imaginary then ` theta` is given by

Options

A. `n pi + (pi)/( 4)`
B. ` n pi - (pi)/( 4)`
C. ` 2 n pi`
D. ` 2 n pi + (pi)/( 4)`

Answer

To determine the values of \( \theta \) for which the expression

\[ \frac{\tan \theta - i \left( \sin \frac{\theta}{2} + \cos \frac{\theta}{2} \right)}{1 + 2i \sin \frac{\theta}{2}} \]

is purely imaginary, we need to ensure that the real part of the expression equals zero.

### Step-by-Step Solution

1. **Identify the expression**:
We start with the expression:
\[ z = \frac{\tan \theta - i \left( \sin \frac{\theta}{2} + \cos \frac{\theta}{2} \right)}{1 + 2i \sin \frac{\theta}{2}} \]

2. **Multiply numerator and denominator by the conjugate of the denominator**:
The conjugate of the denominator \( 1 + 2i \sin \frac{\theta}{2} \) is \( 1 - 2i \sin \frac{\theta}{2} \). Therefore, we multiply both the numerator and denominator by this conjugate:
\[ z = \frac{(\tan \theta - i \left( \sin \frac{\theta}{2} + \cos \frac{\theta}{2} \right))(1 - 2i \sin \frac{\theta}{2})}{(1 + 2i \sin \frac{\theta}{2})(1 - 2i \sin \frac{\theta}{2})} \]

3. **Simplify the denominator**:
The denominator simplifies as follows:
\[ (1 + 2i \sin \frac{\theta}{2})(1 - 2i \sin \frac{\theta}{2}) = 1 - (2i \sin \frac{\theta}{2})^2 = 1 + 4 \sin^2 \frac{\theta}{2} \]

4. **Expand the numerator**:
Now we expand the numerator:
\[ (\tan \theta)(1) - (\tan \theta)(2i \sin \frac{\theta}{2}) - i\left( \sin \frac{\theta}{2} + \cos \frac{\theta}{2} \right)(1) + i\left( \sin \frac{\theta}{2} + \cos \frac{\theta}{2} \right)(2i \sin \frac{\theta}{2}) \]
This simplifies to:
\[ \tan \theta - 2i \tan \theta \sin \frac{\theta}{2} - i \left( \sin \frac{\theta}{2} + \cos \frac{\theta}{2} \right) - 2 \sin^2 \frac{\theta}{2} - 2i \sin \frac{\theta}{2} \cos \frac{\theta}{2} \]

5. **Combine real and imaginary parts**:
The real part of the numerator is:
\[ \tan \theta - 2 \sin^2 \frac{\theta}{2} \]
The imaginary part is:
\[ -2 \tan \theta \sin \frac{\theta}{2} - \left( \sin \frac{\theta}{2} + \cos \frac{\theta}{2} \right) - 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \]

6. **Set the real part to zero**:
For \( z \) to be purely imaginary, we set the real part to zero:
\[ \tan \theta - 2 \sin^2 \frac{\theta}{2} = 0 \]
This gives:
\[ \tan \theta = 2 \sin^2 \frac{\theta}{2} \]

7. **Use the identity**:
Recall that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \):
\[ \frac{\sin \theta}{\cos \theta} = 2 \sin^2 \frac{\theta}{2} \]

8. **Use double angle identities**:
Substitute \( \sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \) into the equation:
\[ \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{\cos \theta} = 2 \sin^2 \frac{\theta}{2} \]
This simplifies to:
\[ \frac{\cos \frac{\theta}{2}}{\cos \theta} = 1 \]
Which implies:
\[ \cos \theta = \cos \frac{\theta}{2} \]

9. **Solve for \( \theta \)**:
The solutions to \( \cos \theta = \cos \frac{\theta}{2} \) yield:
\[ \theta = 2n\pi \quad \text{or} \quad \theta = n\pi + \frac{\pi}{4} \]

### Final Answer
Thus, the values of \( \theta \) for which the expression is purely imaginary are:
\[ \theta = 2n\pi \quad \text{and} \quad \theta = n\pi + \frac{\pi}{4} \]