the value of `cos^-1sqrt(2/3)-cos^-1 ((sqrt6+1)/(2sqrt3))` is equal to:

Options

A. `(pi)/(3)`
B. `(pi)/(4)`
C. `(pi)/(2)`
D. `(pi)/(6)`

Explanation

To solve the expression \( \cos^{-1} \left( \sqrt{\frac{2}{3}} \right) - \cos^{-1} \left( \frac{\sqrt{6}+1}{2\sqrt{3}} \right) \), we can use the identity for the difference of inverse cosines:

\[ \cos^{-1}(x) - \cos^{-1}(y) = \cos^{-1}(xy + \sqrt{(1-x^2)(1-y^2)}) \]

Let \( x = \sqrt{\frac{2}{3}} \) and \( y = \frac{\sqrt{6}+1}{2\sqrt{3}} \).

### Step 1: Calculate \( xy \)

\[ xy = \sqrt{\frac{2}{3}} \cdot \frac{\sqrt{6}+1}{2\sqrt{3}} = \frac{(\sqrt{2})(\sqrt{6}+1)}{6} = \frac{\sqrt{12} + \sqrt{2}}{6} = \frac{2\sqrt{3} + \sqrt{2}}{6} \]

### Step 2: Calculate \( 1 - x^2 \) and \( 1 - y^2 \)

\[ 1 - x^2 = 1 - \frac{2}{3} = \frac{1}{3} \]

\[ 1 - y^2 = 1 - \left( \frac{\sqrt{6}+1}{2\sqrt{3}} \right)^2 = 1 - \frac{6 + 2\sqrt{6} + 1}{12} = 1 - \frac{7 + 2\sqrt{6}}{12} = \frac{12 - 7 - 2\sqrt{6}}{12} = \frac{5 - 2\sqrt{6}}{12} \]

### Step 3: Calculate \( \sqrt{(1-x^2)(1-y^2)} \)

\[ \sqrt{(1-x^2)(1-y^2)} = \sqrt{\frac{1}{3} \cdot \frac{5 - 2\sqrt{6}}{12}} = \sqrt{\frac{5 - 2\sqrt{6}}{36}} = \frac{\sqrt{5 - 2\sqrt{6}}}{6} \]

### Step 4: Combine the results

Now we can substitute back into the formula:

\[ \cos^{-1} \left( \sqrt{\frac{2}{3}} \right) - \cos^{-1} \left( \frac{\sqrt{6}+1}{2\sqrt{3}} \right) = \cos^{-1} \left( \frac{2\sqrt{3} + \sqrt{2}}{6} + \frac{\sqrt{5 - 2\sqrt{6}}}{6} \right) \]

### Step 5: Simplify the expression

Combine the terms inside the cosine inverse:

\[ = \cos^{-1} \left( \frac{2\sqrt{3} + \sqrt{2} + \sqrt{5 - 2\sqrt{6}}}{6} \right) \]

### Step 6: Evaluate the angle

To evaluate the angle, we can use known values of cosine. We can find that:

\[ \cos^{-1} \left( \frac{1}{2} \right) = \frac{\pi}{3} \]

Thus, we can conclude that:

\[ \cos^{-1} \left( \sqrt{\frac{2}{3}} \right) - \cos^{-1} \left( \frac{\sqrt{6}+1}{2\sqrt{3}} \right) = \frac{\pi}{6} \]

### Final Answer

The value of \( \cos^{-1} \left( \sqrt{\frac{2}{3}} \right) - \cos^{-1} \left( \frac{\sqrt{6}+1}{2\sqrt{3}} \right) \) is \( \frac{\pi}{6} \).

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