The value of `tan[(1)/(2)cos^(-1)((sqrt(5))/(3))]` is :

Options

A. `(3+sqrt(5))/(2)`
B. `3+sqrt(5)`
C. `(3-sqrt(5))/(2)`
D. None of these

Answer

To solve the problem \( \tan\left(\frac{1}{2} \cos^{-1}\left(\frac{\sqrt{5}}{3}\right)\right) \), we can follow these steps:

### Step 1: Let \( \theta = \cos^{-1}\left(\frac{\sqrt{5}}{3}\right) \)
This means that \( \cos(\theta) = \frac{\sqrt{5}}{3} \).

### Step 2: Use the half-angle identity for tangent
We know that:
\[ \tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}} \]
So, we can substitute \( \cos(\theta) \) into this identity.

### Step 3: Substitute \( \cos(\theta) \) into the half-angle identity
Substituting \( \cos(\theta) = \frac{\sqrt{5}}{3} \):
\[ \tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \frac{\sqrt{5}}{3}}{1 + \frac{\sqrt{5}}{3}}} \]

### Step 4: Simplify the expression
First, simplify the numerator:
\[ 1 - \frac{\sqrt{5}}{3} = \frac{3 - \sqrt{5}}{3} \]
Now for the denominator:
\[ 1 + \frac{\sqrt{5}}{3} = \frac{3 + \sqrt{5}}{3} \]
Now substituting these back into the tangent expression:
\[ \tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{\frac{3 - \sqrt{5}}{3}}{\frac{3 + \sqrt{5}}{3}}} = \sqrt{\frac{3 - \sqrt{5}}{3 + \sqrt{5}}} \]

### Step 5: Rationalize the denominator
To rationalize the denominator, multiply the numerator and denominator by \( 3 - \sqrt{5} \):
\[ \tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{(3 - \sqrt{5})^2}{(3 + \sqrt{5})(3 - \sqrt{5})}} \]
Calculating the denominator:
\[ (3 + \sqrt{5})(3 - \sqrt{5}) = 9 - 5 = 4 \]
Calculating the numerator:
\[ (3 - \sqrt{5})^2 = 9 - 6\sqrt{5} + 5 = 14 - 6\sqrt{5} \]
Thus, we have:
\[ \tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{14 - 6\sqrt{5}}{4}} = \frac{\sqrt{14 - 6\sqrt{5}}}{2} \]

### Final Answer
The value of \( \tan\left(\frac{1}{2} \cos^{-1}\left(\frac{\sqrt{5}}{3}\right)\right) \) is:
\[ \frac{\sqrt{14 - 6\sqrt{5}}}{2} \]