The value of `cos^(-1)(cos(2tan^(-1)((sqrt(3)+1)/(sqrt(4-2sqrt(3))))))`is

Options

A. `(pi)/6`
B. `(5pi)/6`
C. `(5pi)/12`
D. `(pi)/12`

Explanation

To solve the problem \( \cos^{-1}(\cos(2\tan^{-1}(\frac{\sqrt{3}+1}{\sqrt{4-2\sqrt{3}}}))) \), we can follow these steps:

### Step 1: Simplify the Denominator
First, we simplify the denominator \( \sqrt{4 - 2\sqrt{3}} \).

\[ 4 - 2\sqrt{3} = 1^2 + (\sqrt{3})^2 - 2 \cdot 1 \cdot \sqrt{3} = (\sqrt{3} - 1)^2 \]

Thus,

\[ \sqrt{4 - 2\sqrt{3}} = \sqrt{(\sqrt{3} - 1)^2} = \sqrt{3} - 1 \]

### Step 2: Substitute Back into the Expression
Now, substitute this back into the expression:

\[ \tan^{-1}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right) \]

### Step 3: Simplify the Fraction
Next, we simplify the fraction \( \frac{\sqrt{3}+1}{\sqrt{3}-1} \) by multiplying the numerator and denominator by \( \sqrt{3}+1 \):

\[ \frac{(\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{(\sqrt{3}+1)^2}{3-1} = \frac{3 + 2\sqrt{3} + 1}{2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \]

### Step 4: Find the Angle
Now we need to find the angle whose tangent is \( 2 + \sqrt{3} \). We know:

\[ \tan\left(\frac{\pi}{4} + \frac{\pi}{6}\right) = \frac{\tan\frac{\pi}{4} + \tan\frac{\pi}{6}}{1 - \tan\frac{\pi}{4} \tan\frac{\pi}{6}} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1} \]

Thus,

\[ \tan^{-1}(2+\sqrt{3}) = \frac{\pi}{4} + \frac{\pi}{6} = \frac{3\pi}{12} + \frac{2\pi}{12} = \frac{5\pi}{12} \]

### Step 5: Calculate \( 2\tan^{-1}(2+\sqrt{3}) \)
Now we calculate:

\[ 2\tan^{-1}(2+\sqrt{3}) = 2 \cdot \frac{5\pi}{12} = \frac{10\pi}{12} = \frac{5\pi}{6} \]

### Step 6: Find \( \cos^{-1}(\cos(\frac{5\pi}{6})) \)
Finally, we find:

\[ \cos^{-1}(\cos(\frac{5\pi}{6})) = \frac{5\pi}{6} \]

### Conclusion
Thus, the value of \( \cos^{-1}(\cos(2\tan^{-1}(\frac{\sqrt{3}+1}{\sqrt{4-2\sqrt{3}}}))) \) is

\[ \boxed{\frac{5\pi}{6}} \]