The value of `tan(1/2 cos^(-1)(2/(sqrt(5))))` is

Options

A. `2+sqrt(5)`
B. `sqrt(5) -2`
C. `(sqrt(5)+2)/(2)`
D. `5+sqrt(2)`

Explanation

To find the value of \( \tan\left(\frac{1}{2} \cos^{-1}\left(\frac{2}{\sqrt{5}}\right)\right) \), we will follow these steps:

### Step 1: Set up the equation
Let
\[ \theta = \frac{1}{2} \cos^{-1}\left(\frac{2}{\sqrt{5}}\right) \]
Then we need to find \( \tan(\theta) \).

### Step 2: Express \( \cos(2\theta) \)
Using the double angle identity for cosine, we have:
\[ \cos(2\theta) = \frac{2}{\sqrt{5}} \]
From the double angle formula, we know:
\[ \cos(2\theta) = 2\cos^2(\theta) - 1 \]
Setting these equal gives:
\[ 2\cos^2(\theta) - 1 = \frac{2}{\sqrt{5}} \]

### Step 3: Solve for \( \cos^2(\theta) \)
Rearranging the equation:
\[ 2\cos^2(\theta) = \frac{2}{\sqrt{5}} + 1 \]
\[ 2\cos^2(\theta) = \frac{2 + \sqrt{5}}{\sqrt{5}} \]
Dividing both sides by 2:
\[ \cos^2(\theta) = \frac{2 + \sqrt{5}}{2\sqrt{5}} \]

### Step 4: Find \( \sin^2(\theta) \)
Using the Pythagorean identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ \sin^2(\theta) = 1 - \cos^2(\theta) \]
Substituting the value of \( \cos^2(\theta) \):
\[ \sin^2(\theta) = 1 - \frac{2 + \sqrt{5}}{2\sqrt{5}} = \frac{2\sqrt{5} - (2 + \sqrt{5})}{2\sqrt{5}} = \frac{2\sqrt{5} - 2 - \sqrt{5}}{2\sqrt{5}} = \frac{\sqrt{5} - 2}{2\sqrt{5}} \]

### Step 5: Calculate \( \tan(\theta) \)
Now we can find \( \tan(\theta) \):
\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\sqrt{\sin^2(\theta)}}{\sqrt{\cos^2(\theta)}} \]
Using the values we found:
\[ \tan(\theta) = \frac{\sqrt{\frac{\sqrt{5} - 2}{2\sqrt{5}}}}{\sqrt{\frac{2 + \sqrt{5}}{2\sqrt{5}}}} = \frac{\sqrt{\sqrt{5} - 2}}{\sqrt{2 + \sqrt{5}}} \]

### Step 6: Rationalize the denominator
To rationalize the denominator:
\[ \tan(\theta) = \frac{\sqrt{\sqrt{5} - 2}}{\sqrt{2 + \sqrt{5}}} \cdot \frac{\sqrt{2 - \sqrt{5}}}{\sqrt{2 - \sqrt{5}}} = \frac{\sqrt{(\sqrt{5} - 2)(2 - \sqrt{5})}}{2 - 5} = \frac{\sqrt{5 - 2\sqrt{5} - 4}}{-3} = \frac{\sqrt{1 - 2\sqrt{5}}}{-3} \]

### Final Result
Thus, the value of \( \tan\left(\frac{1}{2} \cos^{-1}\left(\frac{2}{\sqrt{5}}\right)\right) \) simplifies to:
\[ \boxed{\sqrt{5} - 2} \]