Consider the complex number `z = (1 - isin theta)//(1+ icos theta)`.
The value of `theta` for which z is unimodular give by

Options

A. `npi pm (pi)/(6), n in I`
B. `npi pm (pi)/(3), n in I`
C. `npi pm (pi)/(4), n in I`
D. no real values of `theta`

Explanation

To determine the value of \( \theta \) for which the complex number \( z = \frac{1 - i \sin \theta}{1 + i \cos \theta} \) is unimodular, we need to find when the modulus of \( z \) is equal to 1.

### Step-by-Step Solution:

1. **Understanding Unimodular Condition**:
A complex number \( z \) is said to be unimodular if \( |z| = 1 \).

2. **Expressing Modulus**:
Given \( z = \frac{1 - i \sin \theta}{1 + i \cos \theta} \), we can express its modulus as:
\[ |z| = \frac{|1 - i \sin \theta|}{|1 + i \cos \theta|} \]

3. **Calculating Modulus of Numerator**:
The modulus of the numerator \( |1 - i \sin \theta| \) is calculated as:
\[ |1 - i \sin \theta| = \sqrt{(1)^2 + (-\sin \theta)^2} = \sqrt{1 + \sin^2 \theta} \]

4. **Calculating Modulus of Denominator**:
The modulus of the denominator \( |1 + i \cos \theta| \) is calculated as:
\[ |1 + i \cos \theta| = \sqrt{(1)^2 + (\cos \theta)^2} = \sqrt{1 + \cos^2 \theta} \]

5. **Setting Up the Equation**:
For \( z \) to be unimodular, we need:
\[ |z| = \frac{\sqrt{1 + \sin^2 \theta}}{\sqrt{1 + \cos^2 \theta}} = 1 \]

6. **Cross-Multiplying**:
Squaring both sides gives:
\[ 1 + \sin^2 \theta = 1 + \cos^2 \theta \]
Simplifying this leads to:
\[ \sin^2 \theta = \cos^2 \theta \]

7. **Using Trigonometric Identity**:
We can rewrite \( \sin^2 \theta = \cos^2 \theta \) as:
\[ \tan^2 \theta = 1 \]

8. **Finding Angles**:
The solutions to \( \tan^2 \theta = 1 \) are:
\[ \theta = n\pi \pm \frac{\pi}{4}, \quad n \in \mathbb{Z} \]

### Final Answer:
Thus, the values of \( \theta \) for which \( z \) is unimodular are given by:
\[ \theta = n\pi + \frac{\pi}{4} \quad \text{or} \quad \theta = n\pi - \frac{\pi}{4}, \quad n \in \mathbb{Z} \]