If `theta` is the angle between unit vectors `vecA` and `vecB`, then `((1-vecA.vecB))/(1+vecA.vecB))` is equal to

Options

A. `tan^(2)(theta//2)`
B. `sin^(2)(theta//2)`
C. `cot^(2)(theta//2)`
D. `cos^(2)(theta//2)`

Explanation

To solve the problem, we need to evaluate the expression \(\frac{1 - \vec{A} \cdot \vec{B}}{1 + \vec{A} \cdot \vec{B}}\) given that \(\vec{A}\) and \(\vec{B}\) are unit vectors and \(\theta\) is the angle between them.

### Step-by-Step Solution:

1. **Understanding the Dot Product**:
Since \(\vec{A}\) and \(\vec{B}\) are unit vectors, the dot product \(\vec{A} \cdot \vec{B}\) can be expressed as:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta = 1 \cdot 1 \cdot \cos \theta = \cos \theta \]

2. **Substituting the Dot Product**:
We substitute \(\vec{A} \cdot \vec{B}\) in the expression:
\[ \frac{1 - \vec{A} \cdot \vec{B}}{1 + \vec{A} \cdot \vec{B}} = \frac{1 - \cos \theta}{1 + \cos \theta} \]

3. **Using Trigonometric Identities**:
We can use the trigonometric identities:
\[ 1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right) \]
\[ 1 + \cos \theta = 2 \cos^2\left(\frac{\theta}{2}\right) \]
Substituting these identities into our expression gives:
\[ \frac{1 - \cos \theta}{1 + \cos \theta} = \frac{2 \sin^2\left(\frac{\theta}{2}\right)}{2 \cos^2\left(\frac{\theta}{2}\right)} \]

4. **Simplifying the Expression**:
The \(2\)s in the numerator and denominator cancel out:
\[ \frac{\sin^2\left(\frac{\theta}{2}\right)}{\cos^2\left(\frac{\theta}{2}\right)} = \tan^2\left(\frac{\theta}{2}\right) \]

5. **Final Result**:
Therefore, the expression simplifies to:
\[ \frac{1 - \vec{A} \cdot \vec{B}}{1 + \vec{A} \cdot \vec{B}} = \tan^2\left(\frac{\theta}{2}\right) \]

### Conclusion:
The final result is:
\[ \frac{1 - \vec{A} \cdot \vec{B}}{1 + \vec{A} \cdot \vec{B}} = \tan^2\left(\frac{\theta}{2}\right) \]