if `(tanalpha-i(sin ""(alpha)/(2)+cos ""(alpha)/(2)))/(1+2 i sin ""(alpha)/(2))` is purely imaginary then `alpha` is given by -

Options

A. `npi+(pi)/(4)`
B. `npi-(pi)/(4)`
C. `(2n+1)pi`
D. `2npi+(pi)/(4)`

Answer

To solve the problem, we need to determine the values of \( \alpha \) for which the expression

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

is purely imaginary. This means that the real part of this expression must be equal to zero.

### Step-by-Step Solution:

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

2. **Rationalize the Denominator**:
To eliminate the imaginary part in the denominator, we multiply the numerator and denominator by the conjugate of the denominator:
\[ z = \frac{\left(\tan \alpha - i\left(\frac{\sin \alpha}{2} + \frac{\cos \alpha}{2}\right)\right) \cdot \left(1 - 2i\frac{\sin \alpha}{2}\right)}{\left(1 + 2i\frac{\sin \alpha}{2}\right) \cdot \left(1 - 2i\frac{\sin \alpha}{2}\right)} \]

3. **Calculate the Denominator**:
The denominator simplifies as follows:
\[ (1 + 2i\frac{\sin \alpha}{2})(1 - 2i\frac{\sin \alpha}{2}) = 1^2 - (2i\frac{\sin \alpha}{2})^2 = 1 + 4\left(\frac{\sin^2 \alpha}{4}\right) = 1 + \sin^2 \alpha \]

4. **Calculate the Numerator**:
Expanding the numerator:
\[ \tan \alpha - i\left(\frac{\sin \alpha}{2} + \frac{\cos \alpha}{2}\right) - 2i \tan \alpha \frac{\sin \alpha}{2} + 2\left(\frac{\sin \alpha}{2} + \frac{\cos \alpha}{2}\right)\frac{\sin \alpha}{2} \]
This results in:
\[ \tan \alpha + \left(\frac{\sin^2 \alpha}{2} + \frac{\sin \alpha \cos \alpha}{2}\right) + i\left(-\frac{\sin \alpha}{2} + 2\tan \alpha \frac{\sin \alpha}{2}\right) \]

5. **Separate Real and Imaginary Parts**:
The real part \( R \) and imaginary part \( I \) of \( z \) can be expressed as:
\[ R = \frac{\tan \alpha + \frac{\sin^2 \alpha}{2} + \frac{\sin \alpha \cos \alpha}{2}}{1 + \sin^2 \alpha} \]
\[ I = \frac{-\frac{\sin \alpha}{2} + 2\tan \alpha \frac{\sin \alpha}{2}}{1 + \sin^2 \alpha} \]

6. **Set the Real Part to Zero**:
For \( z \) to be purely imaginary, we set the real part \( R \) to zero:
\[ \tan \alpha + \frac{\sin^2 \alpha}{2} + \frac{\sin \alpha \cos \alpha}{2} = 0 \]

7. **Substitute \( \tan \alpha \)**:
Recall that \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \):
\[ \frac{\sin \alpha}{\cos \alpha} + \frac{\sin^2 \alpha}{2} + \frac{\sin \alpha \cos \alpha}{2} = 0 \]

8. **Multiply Through by \( 2\cos \alpha \)**:
This gives:
\[ 2\sin \alpha + \sin^2 \alpha \cos \alpha + \sin \alpha \cos^2 \alpha = 0 \]

9. **Factor Out \( \sin \alpha \)**:
\[ \sin \alpha (2 + \sin \alpha \cos \alpha + \cos^2 \alpha) = 0 \]

10. **Solve for \( \alpha \)**:
- From \( \sin \alpha = 0 \), we have \( \alpha = n\pi \).
- From \( 2 + \sin \alpha \cos \alpha + \cos^2 \alpha = 0 \), we can solve for other values of \( \alpha \).

### Final Values of \( \alpha \):
The values of \( \alpha \) that satisfy the original condition are:
\[ \alpha = n\pi + \frac{\pi}{4} \quad \text{or} \quad \alpha = 2n\pi \]