Find the value of `(1-tan^(2)22(1^(@))/(2))/(1+tan^(2)22(1^(@))/(2))`.

Options

A. `(1)/(sqrt(2))`
B. 1
C. `(1)/(2)`
D. `sqrt(3)/(2)`

Answer

To solve the expression \(\frac{1 - \tan^2(22.5^\circ)}{1 + \tan^2(22.5^\circ)}\), we can use the trigonometric identity for cosine of double angles.

### Step-by-Step Solution:

1. **Identify the expression**: We start with the expression \(\frac{1 - \tan^2(22.5^\circ)}{1 + \tan^2(22.5^\circ)}\).

2. **Use the identity for cosine**: We know that:
\[ \cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} \]
Here, we can let \(\theta = 22.5^\circ\).

3. **Substitute for \(\theta\)**: Thus, we can rewrite our expression as:
\[ \frac{1 - \tan^2(22.5^\circ)}{1 + \tan^2(22.5^\circ)} = \cos(2 \times 22.5^\circ) \]

4. **Calculate \(2\theta\)**: Now, calculate \(2 \times 22.5^\circ\):
\[ 2 \times 22.5^\circ = 45^\circ \]

5. **Find \(\cos(45^\circ)\)**: We know that:
\[ \cos(45^\circ) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

6. **Final Result**: Therefore, the value of the original expression is:
\[ \frac{1 - \tan^2(22.5^\circ)}{1 + \tan^2(22.5^\circ)} = \cos(45^\circ) = \frac{\sqrt{2}}{2} \]

### Final Answer:
\[ \frac{\sqrt{2}}{2} \]